Scalar or Dot product of two vectors and its applications Questions in English

(B) Given $(\vec{a}+3 \vec{b}) \perp(7 \vec{a}-5 \vec{b})$,their dot product is zero:
$(\vec{a}+3 \vec{b}) \cdot(7 \vec{a}-5 \vec{b}) = 7|\vec{a}|^2 - 15|\vec{b}|^2 + 16(\vec{a} \cdot \vec{b}) = 0 \quad \dots(1)$
Given $(\vec{a}-4 \vec{b}) \perp(7 \vec{a}-2 \vec{b})$,their dot product is zero:
$(\vec{a}-4 \vec{b}) \cdot(7 \vec{a}-2 \vec{b}) = 7|\vec{a}|^2 + 8|\vec{b}|^2 - 30(\vec{a} \cdot \vec{b}) = 0 \quad \dots(2)$
Subtracting equation $(2)$ from equation $(1)$:
$(7|\vec{a}|^2 - 15|\vec{b}|^2 + 16(\vec{a} \cdot \vec{b})) - (7|\vec{a}|^2 + 8|\vec{b}|^2 - 30(\vec{a} \cdot \vec{b})) = 0$
$-23|\vec{b}|^2 + 46(\vec{a} \cdot \vec{b}) = 0 \implies 46(\vec{a} \cdot \vec{b}) = 23|\vec{b}|^2 \implies \vec{a} \cdot \vec{b} = \frac{1}{2}|\vec{b}|^2$
Substitute $\vec{a} \cdot \vec{b} = \frac{1}{2}|\vec{b}|^2$ into equation $(1)$:
$7|\vec{a}|^2 - 15|\vec{b}|^2 + 16(\frac{1}{2}|\vec{b}|^2) = 0$
$7|\vec{a}|^2 - 15|\vec{b}|^2 + 8|\vec{b}|^2 = 0 \implies 7|\vec{a}|^2 = 7|\vec{b}|^2 \implies |\vec{a}| = |\vec{b}|$
Now,$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{\frac{1}{2}|\vec{b}|^2}{|\vec{b}||\vec{b}|} = \frac{1}{2}$
$\theta = \cos^{-1}(\frac{1}{2}) = 60^{\circ}$

(C) Given $|(\hat{a}+\hat{b})+2(\hat{a} \times \hat{b})|=2$. Since $(\hat{a}+\hat{b}) \perp (\hat{a} \times \hat{b})$,we have $|\hat{a}+\hat{b}|^2 + 4|\hat{a} \times \hat{b}|^2 = 4$.
Using $|\hat{a}+\hat{b}|^2 = 2+2\cos\theta$ and $|\hat{a} \times \hat{b}|^2 = \sin^2\theta = 1-\cos^2\theta$,we get:
$2+2\cos\theta + 4(1-\cos^2\theta) = 4$
$2+2\cos\theta + 4 - 4\cos^2\theta = 4$
$4\cos^2\theta - 2\cos\theta - 2 = 0 \implies 2\cos^2\theta - \cos\theta - 1 = 0$.
$(2\cos\theta+1)(\cos\theta-1) = 0$.
Since $\theta \in (0, \pi)$,$\cos\theta = -\frac{1}{2}$,so $\theta = \frac{2\pi}{3}$.
For $(S_{1})$: $2|\hat{a} \times \hat{b}| = 2\sin(\frac{2\pi}{3}) = 2(\frac{\sqrt{3}}{2}) = \sqrt{3}$.
$|\hat{a}-\hat{b}| = \sqrt{1+1-2\cos(\frac{2\pi}{3})} = \sqrt{2-2(-\frac{1}{2})} = \sqrt{3}$. Thus $(S_{1})$ is true.
For $(S_{2})$: Projection of $\hat{a}$ on $(\hat{a}+\hat{b})$ is $\frac{\hat{a} \cdot (\hat{a}+\hat{b})}{|\hat{a}+\hat{b}|} = \frac{1+\cos\theta}{\sqrt{2+2\cos\theta}} = \frac{1-\frac{1}{2}}{\sqrt{2-1}} = \frac{1/2}{1} = \frac{1}{2}$. Thus $(S_{2})$ is true.

(C) Given that $\hat{a}$ and $\hat{b}$ are unit vectors,so $|\hat{a}| = 1$ and $|\hat{b}| = 1$.
Given $\hat{b} = \vec{c} + 2(\vec{c} \times \hat{a})$.
Taking the squared magnitude on both sides:
$|\hat{b}|^{2} = |\vec{c} + 2(\vec{c} \times \hat{a})|^{2}$
$1 = |\vec{c}|^{2} + 4|\vec{c} \times \hat{a}|^{2} + 4\vec{c} \cdot (\vec{c} \times \hat{a})$
Since $\vec{c} \cdot (\vec{c} \times \hat{a}) = 0$ because the cross product is perpendicular to $\vec{c}$,we have:
$1 = |\vec{c}|^{2} + 4|\vec{c}|^{2} |\hat{a}|^{2} \sin^{2}(\frac{\pi}{12})$
$1 = |\vec{c}|^{2} + 4|\vec{c}|^{2} (1) \sin^{2}(\frac{\pi}{12})$
Using $\sin(\frac{\pi}{12}) = \sin(15^{\circ}) = \frac{\sqrt{6}-\sqrt{2}}{4}$,so $\sin^{2}(\frac{\pi}{12}) = \frac{6+2-2\sqrt{12}}{16} = \frac{8-4\sqrt{3}}{16} = \frac{2-\sqrt{3}}{4}$.
$1 = |\vec{c}|^{2} [1 + 4(\frac{2-\sqrt{3}}{4})] = |\vec{c}|^{2} (1 + 2 - \sqrt{3}) = |\vec{c}|^{2} (3 - \sqrt{3})$.
$|\vec{c}|^{2} = \frac{1}{3-\sqrt{3}} = \frac{3+\sqrt{3}}{9-3} = \frac{3+\sqrt{3}}{6}$.
Therefore,$|6\vec{c}|^{2} = 36|\vec{c}|^{2} = 36 \times \frac{3+\sqrt{3}}{6} = 6(3+\sqrt{3})$.

(C) Given $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ makes equal angles with axes,so $a_{1} = a_{2} = a_{3} = k$. Thus $\vec{a} = k(\hat{i} + \hat{j} + \hat{k})$.
Projection of $\vec{a}$ on $3 \hat{i} + 4 \hat{j}$ is $\frac{\vec{a} \cdot (3 \hat{i} + 4 \hat{j})}{\sqrt{3^2 + 4^2}} = 7$.
$\frac{k(3 + 4)}{5} = 7 \Rightarrow \frac{7k}{5} = 7 \Rightarrow k = 5$. So $\vec{a} = 5(\hat{i} + \hat{j} + \hat{k})$.
Since $\vec{b}$ is obtained by rotating $\vec{a}$ by $90^{\circ}$ and $\vec{a}, \vec{b}, \hat{i}$ are coplanar,$\vec{b}$ lies in the plane of $\vec{a}$ and $\hat{i}$.
Let $\vec{b} = x\vec{a} + y\hat{i}$. Since $|\vec{b}| = |\vec{a}| = 5\sqrt{3}$ and $\vec{a} \cdot \vec{b} = 0$,we find $\vec{b}$ is perpendicular to $\vec{a}$.
Solving for the projection of $\vec{b}$ on $3\hat{i} + 4\hat{j}$,we get the magnitude as $2$.

(A) Given $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$,$\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$.
We know that $|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} |\vec{b}|^{2}$.
Here,$|\vec{a}|^{2} = 1^{2}+(-1)^{2}+2^{2} = 6$.
$|\vec{a} \times \vec{b}|^{2} = 2^{2}+0^{2}+(-1)^{2} = 5$.
Substituting these values,$5+3^{2} = 6 |\vec{b}|^{2} \implies 14 = 6 |\vec{b}|^{2} \implies |\vec{b}|^{2} = \frac{7}{3}$.
Now,$|\vec{a}-\vec{b}|^{2} = |\vec{a}|^{2}+|\vec{b}|^{2}-2(\vec{a} \cdot \vec{b}) = 6+\frac{7}{3}-2(3) = \frac{7}{3}$.
Thus,$|\vec{a}-\vec{b}| = \sqrt{\frac{7}{3}}$.
The projection of $\vec{b}$ on $\vec{a}-\vec{b}$ is given by $\frac{\vec{b} \cdot (\vec{a}-\vec{b})}{|\vec{a}-\vec{b}|}$.
$= \frac{\vec{b} \cdot \vec{a} - |\vec{b}|^{2}}{|\vec{a}-\vec{b}|} = \frac{3 - \frac{7}{3}}{\sqrt{\frac{7}{3}}} = \frac{\frac{2}{3}}{\sqrt{\frac{7}{3}}} = \frac{2}{3} \times \sqrt{\frac{3}{7}} = \frac{2}{\sqrt{21}}$.

(D) In a triangle $ABC$,$\overrightarrow{BC} + \overrightarrow{CA} + \overrightarrow{AB} = \overrightarrow{0}$,so $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0}$.
From $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0}$,we have $\overrightarrow{b} + \overrightarrow{c} = -\overrightarrow{a}$.
Squaring both sides: $|\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{b} \cdot \overrightarrow{c}) = |\overrightarrow{a}|^2$.
Given $|\overrightarrow{a}| = 6\sqrt{2}$,$|\overrightarrow{b}| = 2\sqrt{3}$,and $\overrightarrow{b} \cdot \overrightarrow{c} = 12$,we have $(2\sqrt{3})^2 + |\overrightarrow{c}|^2 + 2(12) = (6\sqrt{2})^2$.
$12 + |\overrightarrow{c}|^2 + 24 = 72 \implies |\overrightarrow{c}|^2 = 36 \implies |\overrightarrow{c}| = 6$.
For $(S1)$: $|(\overrightarrow{a} \times \overrightarrow{b}) + (\overrightarrow{c} \times \overrightarrow{b})| - |\overrightarrow{c}| = |(\overrightarrow{a} + \overrightarrow{c}) \times \overrightarrow{b}| - |\overrightarrow{c}|$.
Since $\overrightarrow{a} + \overrightarrow{c} = -\overrightarrow{b}$,this becomes $|(-\overrightarrow{b}) \times \overrightarrow{b}| - |\overrightarrow{c}| = |\overrightarrow{0}| - 6 = -6$.
Thus,$(S1)$ is false.
For $(S2)$: $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0} \implies \overrightarrow{a} + \overrightarrow{c} = -\overrightarrow{b}$.
$|\overrightarrow{a} + \overrightarrow{c}|^2 = |-\overrightarrow{b}|^2 \implies |\overrightarrow{a}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{c}) = |\overrightarrow{b}|^2$.
$72 + 36 + 2(\overrightarrow{a} \cdot \overrightarrow{c}) = 12 \implies 2(\overrightarrow{a} \cdot \overrightarrow{c}) = -96 \implies \overrightarrow{a} \cdot \overrightarrow{c} = -48$.
Using $\cos(\angle ABC) = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}||\overrightarrow{BC}|} = \frac{(-\overrightarrow{c}) \cdot \overrightarrow{a}}{|-\overrightarrow{c}| |\overrightarrow{a}|} = \frac{-(\overrightarrow{a} \cdot \overrightarrow{c})}{6 \cdot 6\sqrt{2}} = \frac{48}{36\sqrt{2}} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}$.
Since $\frac{2\sqrt{2}}{3} \neq \sqrt{\frac{2}{3}}$,$(S2)$ is false.

(D) Given $\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$ and $\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$.
The cross product is $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & \beta \\ 3 & -5 & 4 \end{vmatrix} = (4+5\beta)\hat{i} + (3\beta-4\alpha)\hat{j} + (-5\alpha-3)\hat{k}$.
Comparing this with $-\hat{i}+9\hat{j}+12\hat{k}$,we get:
$4+5\beta = -1 \Rightarrow 5\beta = -5 \Rightarrow \beta = -1$.
$-5\alpha-3 = 12 \Rightarrow -5\alpha = 15 \Rightarrow \alpha = -3$.
Check: $3\beta-4\alpha = 3(-1)-4(-3) = -3+12 = 9$ (Correct).
So,$\vec{a} = -3\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - 5\hat{j} + 4\hat{k}$.
Then $\vec{b}-2\vec{a} = (3 - 2(-3))\hat{i} + (-5 - 2(1))\hat{j} + (4 - 2(-1))\hat{k} = 9\hat{i} - 7\hat{j} + 6\hat{k}$.
And $\vec{b}+\vec{a} = (3-3)\hat{i} + (-5+1)\hat{j} + (4-1)\hat{k} = -4\hat{j} + 3\hat{k}$.
The projection of $\vec{v}_1 = \vec{b}-2\vec{a}$ on $\vec{v}_2 = \vec{b}+\vec{a}$ is $\frac{\vec{v}_1 \cdot \vec{v}_2}{|\vec{v}_2|}$.
$\vec{v}_1 \cdot \vec{v}_2 = (9)(0) + (-7)(-4) + (6)(3) = 0 + 28 + 18 = 46$.
$|\vec{v}_2| = \sqrt{0^2 + (-4)^2 + 3^2} = \sqrt{16+9} = 5$.
Projection = $\frac{46}{5}$.

(B) For the angle between two vectors $\vec{u}$ and $\vec{v}$ to be acute,their dot product must be positive,i.e.,$\vec{u} \cdot \vec{v} > 0$.
Given $\vec{u} = a(\log_{e} b) \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $\vec{v} = (\log_{e} b) \hat{i} + 2 \hat{j} + 2a(\log_{e} b) \hat{k}$.
Calculating the dot product: $\vec{u} \cdot \vec{v} = a(\log_{e} b)^2 - 12 + 6a(\log_{e} b) > 0$.
Let $t = \log_{e} b$. Since $b > 1$,we have $t > 0$.
The inequality becomes $at^2 + 6at - 12 > 0$ for all $t > 0$.
If $a \le 0$,then for large $t$,$at^2 + 6at - 12$ will be negative,so $a$ must be positive.
For a quadratic $f(t) = at^2 + 6at - 12$ to be positive for all $t > 0$,the vertex must be at $t = -\frac{6a}{2a} = -3$. Since the vertex is at $t = -3$ (which is outside the domain $t > 0$) and the parabola opens upward $(a > 0)$,the minimum value for $t > 0$ occurs as $t \to 0^+$.
As $t \to 0^+$,$f(t) \to -12$,which is not greater than $0$.
Thus,there is no value of $a$ that satisfies the condition for all $t > 0$.
Therefore,$S = \phi$.

(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are coplanar and the angle between any two is the same,the angle $\theta = \frac{2\pi}{3} = 120^\circ$.
Let $a = |\vec{a}|, b = |\vec{b}|, c = |\vec{c}|$. We are given $abc = 14$.
Using the vector identity $(\vec{u} \times \vec{v}) \cdot (\vec{w} \times \vec{z}) = (\vec{u} \cdot \vec{w})(\vec{v} \cdot \vec{z}) - (\vec{u} \cdot \vec{z})(\vec{v} \cdot \vec{w})$:
$(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{b}) = (a b \cos 120^\circ)(b c \cos 120^\circ) - (a c \cos 120^\circ)(b^2) = (a b \cdot -\frac{1}{2})(b c \cdot -\frac{1}{2}) - (a c \cdot -\frac{1}{2})(b^2) = \frac{1}{4} a b^2 c + \frac{1}{2} a b^2 c = \frac{3}{4} a b^2 c$.
Similarly,$(\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) = \frac{3}{4} a b c^2$ and $(\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = \frac{3}{4} a^2 b c$.
Summing these,we get $\frac{3}{4} a b c (a + b + c) = 168$.
Substituting $abc = 14$: $\frac{3}{4} \cdot 14 \cdot (a + b + c) = 168$.
$\frac{21}{2} (a + b + c) = 168 \implies a + b + c = 168 \cdot \frac{2}{21} = 8 \cdot 2 = 16$.

(A) Given $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}$ and $\vec{a} \cdot \vec{b}=3$.
Expanding the left side,we have $|\vec{a}|^{2}+|\vec{b}|^{2}+2(\vec{a} \cdot \vec{b})=|\vec{a}|^{2}+2|\vec{b}|^{2}$.
Subtracting $|\vec{a}|^{2}$ from both sides,we get $|\vec{b}|^{2}=2(\vec{a} \cdot \vec{b})$.
Substituting $\vec{a} \cdot \vec{b}=3$,we find $|\vec{b}|^{2}=2(3)=6$.
Using the Lagrange identity $|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}=75$.
Substituting the known values,$|\vec{a}|^{2}(6)-(3)^{2}=75$.
$6|\vec{a}|^{2}-9=75$.
$6|\vec{a}|^{2}=84$.
$|\vec{a}|^{2}=14$.

(A) Given $\hat{a} \cdot \hat{b} = |\hat{a}||\hat{b}| \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
Let $\vec{u} = \hat{a} + \hat{b}$ and $\vec{v} = \hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})$.
$|\vec{u}|^2 = |\hat{a}|^2 + |\hat{b}|^2 + 2(\hat{a} \cdot \hat{b}) = 1 + 1 + 2(\frac{1}{\sqrt{2}}) = 2 + \sqrt{2}$.
$|\vec{v}|^2 = |\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})|^2 = |\hat{a}|^2 + 4|\hat{b}|^2 + 4|\hat{a} \times \hat{b}|^2 + 4(\hat{a} \cdot \hat{b}) + 0 + 0$.
Since $|\hat{a} \times \hat{b}| = |\hat{a}||\hat{b}| \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,then $|\hat{a} \times \hat{b}|^2 = \frac{1}{2}$.
$|\vec{v}|^2 = 1 + 4(1) + 4(\frac{1}{2}) + 4(\frac{1}{\sqrt{2}}) = 1 + 4 + 2 + 2\sqrt{2} = 7 + 2\sqrt{2}$.
Now,$\vec{u} \cdot \vec{v} = (\hat{a} + \hat{b}) \cdot (\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})) = |\hat{a}|^2 + 2(\hat{a} \cdot \hat{b}) + 0 + (\hat{b} \cdot \hat{a}) + 2|\hat{b}|^2 + 0 = 1 + 2(\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} + 2 = 3 + \frac{3}{\sqrt{2}}$.
$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} = \frac{3 + \frac{3}{\sqrt{2}}}{\sqrt{2+\sqrt{2}}\sqrt{7+2\sqrt{2}}}$.
$\cos^2 \theta = \frac{9(1 + \frac{1}{\sqrt{2}})^2}{(2+\sqrt{2})(7+2\sqrt{2})} = \frac{9(\frac{\sqrt{2}+1}{\sqrt{2}})^2}{14 + 4\sqrt{2} + 7\sqrt{2} + 4} = \frac{9(\frac{3+2\sqrt{2}}{2})}{18 + 11\sqrt{2}} = \frac{9(3+2\sqrt{2})}{2(18+11\sqrt{2})}$.
Multiplying by $164$: $164 \cos^2 \theta = 164 \times \frac{9(3+2\sqrt{2})}{2(18+11\sqrt{2})} = 82 \times \frac{9(3+2\sqrt{2})}{18+11\sqrt{2}} \times \frac{18-11\sqrt{2}}{18-11\sqrt{2}} = 738 \times \frac{54 - 33\sqrt{2} + 36\sqrt{2} - 44}{324 - 242} = 738 \times \frac{10 + 3\sqrt{2}}{82} = 9(10 + 3\sqrt{2}) = 90 + 27\sqrt{2}$.

(A) The equation $\vec{r} \cdot \vec{a} = 5$ represents a plane $x + y + z = 5$.
The equation $|\vec{r} - \vec{b}| + |\vec{r} - \vec{c}| = 4$ represents an ellipse in space with foci at $\vec{b}$ and $\vec{c}$,provided the plane contains the foci.
First,check if $\vec{b}$ and $\vec{c}$ lie on the plane $x + y + z = 5$:
For $\vec{b}(2, 2, 1)$,$2 + 2 + 1 = 5$ (True).
For $\vec{c}(5, 1, -1)$,$5 + 1 - 1 = 5$ (True).
Since both foci lie on the plane,the intersection is an ellipse in the plane.
The length of the major axis is $2a = 4$,so $a = 2$.
The distance between the foci is $2ae = |\vec{b} - \vec{c}| = |(2-5)\hat{i} + (2-1)\hat{j} + (1-(-1))\hat{k}| = |-3\hat{i} + \hat{j} + 2\hat{k}| = \sqrt{(-3)^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$.
Thus,$2ae = \sqrt{14} \Rightarrow 4e = \sqrt{14} \Rightarrow e = \frac{\sqrt{14}}{4}$.
Using $b^2 = a^2(1 - e^2)$,we get $b^2 = 4(1 - \frac{14}{16}) = 4(\frac{2}{16}) = \frac{8}{16} = \frac{1}{2}$,so $b = \frac{1}{\sqrt{2}}$.
The area of the ellipse is $\pi ab = \pi(2)(\frac{1}{\sqrt{2}}) = \sqrt{2}\pi \approx 1.414 \times 3.14159 \approx 4.44$.
The closest integer is $4$.

(C) Let the position vectors of the vertices of $\triangle ABC$ be $\vec{a}, \vec{b},$ and $\vec{c}$ respectively.
Let the circumcentre $O$ of $\triangle ABC$ be the origin,so $\vec{O} = \vec{0}$.
The position vector of the orthocentre $H$ is given by $\vec{H} = \vec{a} + \vec{b} + \vec{c}$.
Since $N$ is the mid-point of $OH$,its position vector $\vec{n}$ is given by $\vec{n} = \frac{\vec{O} + \vec{H}}{2} = \frac{\vec{a} + \vec{b} + \vec{c}}{2}$.
We need to find the sum $\overrightarrow{NA} + \overrightarrow{NB} + \overrightarrow{NC}$.
This is equal to $(\vec{a} - \vec{n}) + (\vec{b} - \vec{n}) + (\vec{c} - \vec{n})$.
$= (\vec{a} + \vec{b} + \vec{c}) - 3\vec{n}$.
Substituting $\vec{n} = \frac{\vec{a} + \vec{b} + \vec{c}}{2}$,we get:
$= (\vec{a} + \vec{b} + \vec{c}) - 3 \left( \frac{\vec{a} + \vec{b} + \vec{c}}{2} \right)$.
$= (\vec{a} + \vec{b} + \vec{c}) \left( 1 - \frac{3}{2} \right) = -\frac{1}{2} (\vec{a} + \vec{b} + \vec{c})$.
Since $\vec{H} = \vec{a} + \vec{b} + \vec{c}$,this is equal to $-\frac{1}{2} \vec{H} = -\frac{1}{2} \overrightarrow{OH} = \frac{1}{2} \overrightarrow{HO}$.

(B) Given $\vec{a} = 6 \hat{i} - 3 \hat{j} - 6 \hat{k}$ and $\vec{d} = \hat{i} + \hat{j} + \hat{k}$.
Since $\vec{b}$ is parallel to $\vec{d}$,we can write $\vec{b} = \lambda \vec{d} = \lambda(\hat{i} + \hat{j} + \hat{k})$.
Given $\vec{a} = \vec{b} + \vec{c}$,we have $\vec{c} = \vec{a} - \vec{b} = (6 - \lambda) \hat{i} - (3 + \lambda) \hat{j} - (6 + \lambda) \hat{k}$.
Since $\vec{c}$ is perpendicular to $\vec{d}$,their dot product is zero: $\vec{c} \cdot \vec{d} = 0$.
$(6 - \lambda)(1) + (-3 - \lambda)(1) + (-6 - \lambda)(1) = 0$.
$6 - \lambda - 3 - \lambda - 6 - \lambda = 0$.
$-3 - 3\lambda = 0 \Rightarrow \lambda = -1$.
Substituting $\lambda = -1$ into the expression for $\vec{c}$:
$\vec{c} = (6 - (-1)) \hat{i} - (3 + (-1)) \hat{j} - (6 + (-1)) \hat{k} = 7 \hat{i} - 2 \hat{j} - 5 \hat{k}$.

(C) Let the unit vectors be represented by their angles $\theta_1, \theta_2, \theta_3, \theta_4$ where $\theta_1 \in (0, 90^{\circ})$,$\theta_2 \in (90^{\circ}, 180^{\circ})$,$\theta_3 \in (180^{\circ}, 270^{\circ})$,and $\theta_4 \in (270^{\circ}, 360^{\circ})$.
$(a)$ The sum $v_1 + v_2 + v_3 + v_4$ is not necessarily zero. For example,if the vectors are very close to the positive $X$-axis and positive $Y$-axis,the sum will not be zero.
$(b)$ The sum $v_i + v_j$ is not necessarily in the first quadrant. For instance,if $v_1$ is near $90^{\circ}$ and $v_2$ is near $180^{\circ}$,their sum will have a negative $X$-component.
$(c)$ Consider $v_1$ in the first quadrant and $v_3$ in the third quadrant. The angle between them is $| heta_1 - \theta_3|$. Since $\theta_1 \in (0, 90^{\circ})$ and $\theta_3 \in (180^{\circ}, 270^{\circ})$,the difference $\theta_3 - \theta_1$ lies in $(90^{\circ}, 270^{\circ})$. The dot product $v_1 \cdot v_3 = \cos(\theta_1 - \theta_3)$. Since the angle difference can be greater than $90^{\circ}$,the cosine can be negative. Specifically,if we choose $v_1$ near $45^{\circ}$ and $v_3$ near $225^{\circ}$,the angle is $180^{\circ}$,and the dot product is $\cos(180^{\circ}) = -1 < 0$. Thus,there always exist $i, j$ such that $v_i \cdot v_j < 0$.
$(d)$ This is not necessarily true for all pairs,as shown in $(c)$.

(C) Let $AM = x AB$ and $AN = y AC$. Since $G$ is the centroid,the position vector of $G$ is $\frac{A+B+C}{3}$.
Since $M, G, N$ are collinear,there exists a scalar $k$ such that $\vec{G} = (1-k)\vec{M} + k\vec{N}$.
Using the property of the centroid and the given conditions,it can be shown that $\frac{1}{3x} + \frac{1}{3y} = 1$,or $\frac{1}{x} + \frac{1}{y} = 3$.
The ratio of the areas is $r = \frac{\text{Area}(\triangle AMN)}{\text{Area}(\triangle ABC)} = xy$.
From $\frac{1}{x} + \frac{1}{y} = 3$,we have $y = \frac{x}{3x-1}$.
Thus $r = \frac{x^2}{3x-1}$.
Since $M$ and $N$ are in the interior of the segments,$0 < x < 1$ and $0 < y < 1$.
For $y < 1$,$\frac{x}{3x-1} < 1 \implies x > 1/2$.
Also,the minimum value of $r$ occurs when $x=y=2/3$,giving $r = (2/3)(2/3) = 4/9$.
As $x \to 1/2$,$y \to 1$,so $r \to 1/2$.
Thus,$4/9 \leq r < 1/2$.

(D) Given,$\overrightarrow{PA} + 2\overrightarrow{PB} + 3\overrightarrow{PC} = \vec{0}$.
Let the position vectors of $A, B, C$ and $P$ be $\vec{a}, \vec{b}, \vec{c}$ and $\vec{p}$ respectively.
Then,$(\vec{a} - \vec{p}) + 2(\vec{b} - \vec{p}) + 3(\vec{c} - \vec{p}) = \vec{0}$.
$\vec{a} + 2\vec{b} + 3\vec{c} = 6\vec{p} \implies \vec{p} = \frac{\vec{a} + 2\vec{b} + 3\vec{c}}{6}$.
The area of $\triangle ABC$ is given by $\Delta = \frac{1}{2} |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}|$.
The area of $\triangle APC$ is given by $\Delta_{APC} = \frac{1}{2} |\vec{p} \times \vec{a} + \vec{a} \times \vec{c} + \vec{c} \times \vec{p}|$.
Substituting $\vec{p} = \frac{\vec{a} + 2\vec{b} + 3\vec{c}}{6}$:
$\Delta_{APC} = \frac{1}{2} |\frac{\vec{a} + 2\vec{b} + 3\vec{c}}{6} \times \vec{a} + \vec{a} \times \vec{c} + \vec{c} \times \frac{\vec{a} + 2\vec{b} + 3\vec{c}}{6}|$
$= \frac{1}{12} |\vec{a} \times \vec{a} + 2\vec{b} \times \vec{a} + 3\vec{c} \times \vec{a} + 6(\vec{a} \times \vec{c}) + \vec{c} \times \vec{a} + 2\vec{c} \times \vec{b} + 3\vec{c} \times \vec{c}|$
$= \frac{1}{12} |0 - 2(\vec{a} \times \vec{b}) + 3(\vec{c} \times \vec{a}) - 6(\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) - 2(\vec{b} \times \vec{c}) + 0|$
$= \frac{1}{12} |-2(\vec{a} \times \vec{b}) - 2(\vec{b} \times \vec{c}) - 2(\vec{c} \times \vec{a})| = \frac{1}{6} |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| = \frac{1}{3} \Delta$.
Therefore,$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle APC)} = \frac{\Delta}{\Delta/3} = 3$.

(A) Given $\vec{u}=(1, -1, -2)$,$\vec{v}=(2, 1, -1)$,and $\vec{v} \cdot \vec{w}=2$.
We are given the equation $\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v} \quad \dots(1)$.
Taking the dot product of equation $(1)$ with $\vec{v}$:
$\vec{v} \cdot (\vec{v} \times \vec{w}) = \vec{v} \cdot \vec{u} + \lambda(\vec{v} \cdot \vec{v})$.
Since $\vec{v} \cdot (\vec{v} \times \vec{w}) = 0$,we have $0 = (2 - 1 + 2) + \lambda(2^2 + 1^2 + (-1)^2)$.
$0 = 3 + 6\lambda \implies \lambda = -\frac{1}{2}$.
Now,taking the dot product of equation $(1)$ with $\vec{w}$:
$\vec{w} \cdot (\vec{v} \times \vec{w}) = \vec{w} \cdot \vec{u} + \lambda(\vec{w} \cdot \vec{v})$.
Since $\vec{w} \cdot (\vec{v} \times \vec{w}) = 0$,we have $0 = \vec{u} \cdot \vec{w} + \lambda(2)$.
$\vec{u} \cdot \vec{w} = -2\lambda = -2(-\frac{1}{2}) = 1$.

(B) Let the position vectors of vertices $P, Q, R$ be $\vec{p}, \vec{q}, \vec{r}$ respectively. Without loss of generality,let $\vec{p} = \vec{0}$.
Given $\frac{QA}{AR} = \frac{1}{2}$,point $A$ divides $QR$ in ratio $1:2$. Thus,$\vec{a} = \frac{2\vec{q} + 1\vec{r}}{3}$.
Given $\frac{RB}{BP} = \frac{1}{2}$,point $B$ divides $RP$ in ratio $1:2$. Thus,$\vec{b} = \frac{2\vec{r} + 1\vec{p}}{3} = \frac{2\vec{r}}{3}$.
Given $\frac{PC}{CQ} = \frac{1}{2}$,point $C$ divides $PQ$ in ratio $1:2$. Thus,$\vec{c} = \frac{2\vec{p} + 1\vec{q}}{3} = \frac{\vec{q}}{3}$.
The area of $\triangle PQR$ is given by $\Delta = \frac{1}{2} |\vec{q} \times \vec{r}|$.
The area of $\triangle ABC$ is given by $\Delta' = \frac{1}{2} |(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})|$.
Calculating vectors: $\vec{b} - \vec{a} = \frac{2\vec{r} - 2\vec{q} - \vec{r}}{3} = \frac{\vec{r} - 2\vec{q}}{3}$ and $\vec{c} - \vec{a} = \frac{\vec{q} - 2\vec{q} - \vec{r}}{3} = \frac{-\vec{q} - \vec{r}}{3}$.
$\Delta' = \frac{1}{2} |\frac{1}{9} (\vec{r} - 2\vec{q}) \times (-\vec{q} - \vec{r})| = \frac{1}{18} |-\vec{r} \times \vec{q} - \vec{r} \times \vec{r} + 2\vec{q} \times \vec{q} + 2\vec{q} \times \vec{r}|$.
Since $\vec{x} \times \vec{x} = 0$ and $\vec{r} \times \vec{q} = -\vec{q} \times \vec{r}$,we have $\Delta' = \frac{1}{18} |\vec{q} \times \vec{r} + 2\vec{q} \times \vec{r}| = \frac{1}{18} |3(\vec{q} \times \vec{r})| = \frac{1}{6} |\vec{q} \times \vec{r}|$.
Thus,$\frac{\operatorname{Area}(\triangle PQR)}{\operatorname{Area}(\triangle ABC)} = \frac{\frac{1}{2} |\vec{q} \times \vec{r}|}{\frac{1}{6} |\vec{q} \times \vec{r}|} = 3$.

(C) Let $\vec{\beta}_1 = \lambda \vec{\alpha}$.
Since $\vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2$,we have $\vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1 = \vec{\beta} - \lambda \vec{\alpha}$.
Substituting the vectors,$\vec{\beta}_2 = (\hat{i} + 2\hat{j} - 4\hat{k}) - \lambda(4\hat{i} + 3\hat{j} + 5\hat{k}) = (1 - 4\lambda)\hat{i} + (2 - 3\lambda)\hat{j} - (4 + 5\lambda)\hat{k}$.
Since $\vec{\beta}_2 \perp \vec{\alpha}$,their dot product is zero: $\vec{\beta}_2 \cdot \vec{\alpha} = 0$.
$4(1 - 4\lambda) + 3(2 - 3\lambda) + 5(-4 - 5\lambda) = 0$.
$4 - 16\lambda + 6 - 9\lambda - 20 - 25\lambda = 0$.
$-50\lambda - 10 = 0 \Rightarrow \lambda = -\frac{1}{5}$.
Now,$\vec{\beta}_2 = (1 - 4(-\frac{1}{5}))\hat{i} + (2 - 3(-\frac{1}{5}))\hat{j} - (4 + 5(-\frac{1}{5}))\hat{k} = \frac{9}{5}\hat{i} + \frac{13}{5}\hat{j} - 3\hat{k}$.
Thus,$5\vec{\beta}_2 = 9\hat{i} + 13\hat{j} - 15\hat{k}$.
Finally,$5\vec{\beta}_2 \cdot (\hat{i} + \hat{j} + \hat{k}) = 9(1) + 13(1) - 15(1) = 9 + 13 - 15 = 7$.

(A) Given $\overrightarrow{r} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{0}$.
This can be written as $\overrightarrow{r} \times \overrightarrow{b} - \overrightarrow{c} \times \overrightarrow{b} = \overrightarrow{0}$,which implies $(\overrightarrow{r} - \overrightarrow{c}) \times \overrightarrow{b} = \overrightarrow{0}$.
This means $\overrightarrow{r} - \overrightarrow{c} = \lambda \overrightarrow{b}$ for some scalar $\lambda$,so $\overrightarrow{r} = \overrightarrow{c} + \lambda \overrightarrow{b}$.
Given $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,we substitute $\overrightarrow{r}$:
$(\overrightarrow{c} + \lambda \overrightarrow{b}) \cdot \overrightarrow{a} = 0 \Rightarrow \overrightarrow{c} \cdot \overrightarrow{a} + \lambda (\overrightarrow{b} \cdot \overrightarrow{a}) = 0$.
Calculating the dot products:
$\overrightarrow{c} \cdot \overrightarrow{a} = (7)(1) + (-3)(0) + (4)(2) = 7 + 0 + 8 = 15$.
$\overrightarrow{b} \cdot \overrightarrow{a} = (1)(1) + (1)(0) + (1)(2) = 1 + 0 + 2 = 3$.
Thus,$15 + \lambda(3) = 0 \Rightarrow \lambda = -5$.
Now,$\overrightarrow{r} = \overrightarrow{c} - 5\overrightarrow{b} = (7\hat{i} - 3\hat{j} + 4\hat{k}) - 5(\hat{i} + \hat{j} + \hat{k}) = 2\hat{i} - 8\hat{j} - \hat{k}$.
Finally,$\overrightarrow{r} \cdot \overrightarrow{c} = (2\hat{i} - 8\hat{j} - \hat{k}) \cdot (7\hat{i} - 3\hat{j} + 4\hat{k}) = (2)(7) + (-8)(-3) + (-1)(4) = 14 + 24 - 4 = 34$.

(A) Given $\vec{a}=4 \hat{i}+3 \hat{j}$ and $\vec{b}=3 \hat{i}-4 \hat{j}+5 \hat{k}$.
First,calculate $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 0 \\ 3 & -4 & 5 \end{vmatrix} = \hat{i}(15-0) - \hat{j}(20-0) + \hat{k}(-16-9) = 15 \hat{i} - 20 \hat{j} - 25 \hat{k}$.
Let $\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}$.
From $\vec{c} \cdot (\vec{a} \times \vec{b}) + 25 = 0$,we get $15x - 20y - 25z = -25$,which simplifies to $3x - 4y - 5z = -5$.
From $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$,we get $x + y + z = 4$.
From the projection of $\vec{c}$ on $\vec{a}$ being $1$,$\frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} = 1 \Rightarrow \frac{4x + 3y}{5} = 1 \Rightarrow 4x + 3y = 5$.
Solving the system of equations:
$1) 3x - 4y - 5z = -5$
$2) x + y + z = 4 \Rightarrow 5x + 5y + 5z = 20$
Adding $(1)$ and $(2)$: $8x + y = 15 \Rightarrow y = 15 - 8x$.
Substitute into $4x + 3y = 5$: $4x + 3(15 - 8x) = 5 \Rightarrow 4x + 45 - 24x = 5 \Rightarrow -20x = -40 \Rightarrow x = 2$.
Then $y = 15 - 8(2) = -1$,and $z = 4 - 2 - (-1) = 3$.
So,$\vec{c} = 2 \hat{i} - \hat{j} + 3 \hat{k}$.
The projection of $\vec{c}$ on $\vec{b}$ is $\frac{\vec{c} \cdot \vec{b}}{|\vec{b}|} = \frac{(2)(3) + (-1)(-4) + (3)(5)}{\sqrt{3^2 + (-4)^2 + 5^2}} = \frac{6 + 4 + 15}{\sqrt{9 + 16 + 25}} = \frac{25}{\sqrt{50}} = \frac{25}{5\sqrt{2}} = \frac{5}{\sqrt{2}}$.

(B) Given that $\vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b}$.
We need to find the value of $\vec{b} \cdot \vec{c}$.
$\vec{b} \cdot \vec{c} = \vec{b} \cdot ((2 \vec{a} \times \vec{b}) - 3 \vec{b})$.
Using the distributive property of the dot product,we get:
$\vec{b} \cdot \vec{c} = 2 \vec{b} \cdot (\vec{a} \times \vec{b}) - 3 \vec{b} \cdot \vec{b}$.
Since $\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ because the cross product $\vec{a} \times \vec{b}$ is perpendicular to both $\vec{a}$ and $\vec{b}$,the first term becomes $0$.
Thus,$\vec{b} \cdot \vec{c} = 0 - 3 |\vec{b}|^2$.
Given $|\vec{b}| = 4$,we have $|\vec{b}|^2 = 16$.
Therefore,$\vec{b} \cdot \vec{c} = -3 \times 16 = -48$.

(C) Given $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a} + \vec{b} - \vec{c}|^2$.
Expanding both sides using the property $|\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2\vec{u} \cdot \vec{v}$:
$|\vec{a} + \vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} + \vec{b}) \cdot \vec{c} = |\vec{a} + \vec{b}|^2 + |\vec{c}|^2 - 2(\vec{a} + \vec{b}) \cdot \vec{c}$.
This simplifies to $4(\vec{a} + \vec{b}) \cdot \vec{c} = 0$.
Since $\vec{b} \cdot \vec{c} = 0$,we have $4(\vec{a} \cdot \vec{c}) = 0$,which implies $\vec{a} \cdot \vec{c} = 0$.
Statement $(B)$ claims $\vec{a}$ and $\vec{c}$ are parallel,but $\vec{a} \cdot \vec{c} = 0$ implies they are perpendicular (since $\vec{c} \neq 0$). Thus,$(B)$ is incorrect.
For statement $(A)$,consider $|\overrightarrow{a} + \lambda \overrightarrow{c}|^2 = |\overrightarrow{a}|^2 + \lambda^2 |\overrightarrow{c}|^2 + 2\lambda(\vec{a} \cdot \vec{c})$.
Since $\vec{a} \cdot \vec{c} = 0$,this becomes $|\overrightarrow{a}|^2 + \lambda^2 |\overrightarrow{c}|^2$.
Since $\lambda^2 |\overrightarrow{c}|^2 \geq 0$ for all $\lambda \in R$,it follows that $|\overrightarrow{a} + \lambda \overrightarrow{c}|^2 \geq |\overrightarrow{a}|^2$,which means $|\overrightarrow{a} + \lambda \overrightarrow{c}| \geq |\overrightarrow{a}|$.
Thus,$(A)$ is correct.

(A) Given: $|\vec{a}|=\sqrt{14}$,$|\vec{b}|=\sqrt{6}$,and $|\vec{a} \times \vec{b}|=\sqrt{48}$.
We use the Lagrange's identity for vectors: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$.
Substitute the given values into the identity:
$(\sqrt{48})^2 + (\vec{a} \cdot \vec{b})^2 = (\sqrt{14})^2 \times (\sqrt{6})^2$.
$48 + (\vec{a} \cdot \vec{b})^2 = 14 \times 6$.
$48 + (\vec{a} \cdot \vec{b})^2 = 84$.
$(\vec{a} \cdot \vec{b})^2 = 84 - 48$.
$(\vec{a} \cdot \vec{b})^2 = 36$.

(C) Given $\overrightarrow{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\overrightarrow{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\overrightarrow{c} = 5\hat{i} - 3\hat{j} + 3\hat{k}$.
From $\overrightarrow{r} \times \overrightarrow{b} = \overrightarrow{c} \times \overrightarrow{b}$,we have $(\overrightarrow{r} - \overrightarrow{c}) \times \overrightarrow{b} = 0$.
This implies $\overrightarrow{r} - \overrightarrow{c} = \lambda \overrightarrow{b}$ for some scalar $\lambda$,so $\overrightarrow{r} = \overrightarrow{c} + \lambda \overrightarrow{b}$.
Given $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,we substitute $\overrightarrow{r}$:
$(\overrightarrow{c} + \lambda \overrightarrow{b}) \cdot \overrightarrow{a} = 0 \Rightarrow \overrightarrow{c} \cdot \overrightarrow{a} + \lambda(\overrightarrow{b} \cdot \overrightarrow{a}) = 0$.
Calculate dot products:
$\overrightarrow{c} \cdot \overrightarrow{a} = (5)(1) + (-3)(2) + (3)(3) = 5 - 6 + 9 = 8$.
$\overrightarrow{b} \cdot \overrightarrow{a} = (1)(1) + (-1)(2) + (2)(3) = 1 - 2 + 6 = 5$.
Thus,$8 + 5\lambda = 0 \Rightarrow \lambda = -\frac{8}{5}$.
Now,$\overrightarrow{r} = \overrightarrow{c} - \frac{8}{5}\overrightarrow{b} = (5\hat{i} - 3\hat{j} + 3\hat{k}) - \frac{8}{5}(\hat{i} - \hat{j} + 2\hat{k}) = \frac{1}{5}(25\hat{i} - 15\hat{j} + 15\hat{k} - 8\hat{i} + 8\hat{j} - 16\hat{k}) = \frac{1}{5}(17\hat{i} - 7\hat{j} - \hat{k})$.
$|\overrightarrow{r}|^2 = \frac{1}{25}(17^2 + (-7)^2 + (-1)^2) = \frac{1}{25}(289 + 49 + 1) = \frac{339}{25}$.
Therefore,$25|\overrightarrow{r}|^2 = 339$.

(A) Given $\overrightarrow{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{k}$,and $\overrightarrow{c} = \hat{i} + 2\hat{j} - 3\hat{k}$.
From the condition $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{c} \times \overrightarrow{a}$,we have $(\overrightarrow{r} - \overrightarrow{c}) \times \overrightarrow{a} = 0$.
This implies that $\overrightarrow{r} - \overrightarrow{c}$ is parallel to $\overrightarrow{a}$,so $\overrightarrow{r} = \overrightarrow{c} + \lambda\overrightarrow{a}$ for some scalar $\lambda$.
Given $\overrightarrow{r} \cdot \overrightarrow{b} = 0$,we substitute $\overrightarrow{r}$:
$(\overrightarrow{c} + \lambda\overrightarrow{a}) \cdot \overrightarrow{b} = 0 \Rightarrow \overrightarrow{c} \cdot \overrightarrow{b} + \lambda(\overrightarrow{a} \cdot \overrightarrow{b}) = 0$.
Calculate the dot products:
$\overrightarrow{c} \cdot \overrightarrow{b} = (1)(1) + (2)(0) + (-3)(1) = 1 - 3 = -2$.
$\overrightarrow{a} \cdot \overrightarrow{b} = (2)(1) + (-7)(0) + (5)(1) = 2 + 5 = 7$.
Substituting these values: $-2 + 7\lambda = 0 \Rightarrow \lambda = \frac{2}{7}$.
Now,$\overrightarrow{r} = \overrightarrow{c} + \frac{2}{7}\overrightarrow{a} = (\hat{i} + 2\hat{j} - 3\hat{k}) + \frac{2}{7}(2\hat{i} - 7\hat{j} + 5\hat{k}) = (1 + \frac{4}{7})\hat{i} + (2 - 2)\hat{j} + (-3 + \frac{10}{7})\hat{k} = \frac{11}{7}\hat{i} - \frac{11}{7}\hat{k}$.
Finally,$|\overrightarrow{r}| = \sqrt{(\frac{11}{7})^2 + 0^2 + (-\frac{11}{7})^2} = \sqrt{2 \times (\frac{11}{7})^2} = \frac{11}{7}\sqrt{2}$.

(D) The projection of vector $\vec{a}$ on vector $\vec{b}$ is given by the formula: $\text{Proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
Given $\vec{a} = 5\hat{i} - \hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + 5\hat{k}$.
First,calculate the dot product: $\vec{a} \cdot \vec{b} = (5)(1) + (-1)(3) + (-3)(5) = 5 - 3 - 15 = -13$.
Next,calculate the magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{1^2 + 3^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.
The scalar projection is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{-13}{\sqrt{35}}$.
Since the scalar projection is negative,the direction of the projection vector is opposite to the direction of $\vec{b}$.
Note: The provided options contain a calculation error in the numerator ($17$ instead of $13$). Based on the sign of the dot product,the correct choice is $D$.

(B) Given $\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$,we have $\vec{a} \times (\vec{c} - \vec{b}) = \vec{0}$.
This implies that $(\vec{c} - \vec{b})$ is parallel to $\vec{a}$.
Thus,$\vec{c} - \vec{b} = k \vec{a}$ for some scalar $k$,or $\vec{c} = \vec{b} + k \vec{a}$.
Substituting $\vec{c} = (\alpha + 6k) \hat{i} + (11 + 9k) \hat{j} + (-2 + 12k) \hat{k}$ into $\vec{a} \cdot \vec{c} = -12$:
$6(\alpha + 6k) + 9(11 + 9k) + 12(-2 + 12k) = -12$.
$6\alpha + 36k + 99 + 81k - 24 + 144k = -12 \Rightarrow 6\alpha + 261k = -87$.
Using $\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 5$:
$(\alpha + 6k) - 2(11 + 9k) + (-2 + 12k) = 5$.
$\alpha + 6k - 22 - 18k - 2 + 12k = 5 \Rightarrow \alpha = 29$.
Substituting $\alpha = 29$ into $6(29) + 261k = -87$:
$174 + 261k = -87 \Rightarrow 261k = -261 \Rightarrow k = -1$.
Thus,$\vec{c} = \vec{b} - \vec{a} = (29-6)\hat{i} + (11-9)\hat{j} + (-2-12)\hat{k} = 23\hat{i} + 2\hat{j} - 14\hat{k}$.
Finally,$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 23 + 2 - 14 = 11$.

(A) Let the radius of the circle be $r$. Then $|\vec{u}| = |\vec{v}| = |\vec{OQ}| = r$.
Since $\angle POQ = 90^{\circ}$ and $R$ is the midpoint of arc $PQ$,$\angle POR = \angle ROQ = 45^{\circ}$.
Given $\vec{OQ} = \alpha \vec{u} + \beta \vec{v}$.
Taking dot product with $\vec{u}$:
$\vec{u} \cdot \vec{OQ} = \alpha |\vec{u}|^2 + \beta (\vec{u} \cdot \vec{v})$
Since $\angle POQ = 90^{\circ}$,$\vec{u} \cdot \vec{OQ} = 0$. Also $\vec{u} \cdot \vec{v} = r^2 \cos 45^{\circ} = \frac{r^2}{\sqrt{2}}$.
$0 = \alpha r^2 + \beta \frac{r^2}{\sqrt{2}} \implies \alpha = -\frac{\beta}{\sqrt{2}} \implies \alpha^2 = \frac{\beta^2}{2}$.
Now,$|\vec{OQ}|^2 = r^2 = |\alpha \vec{u} + \beta \vec{v}|^2 = \alpha^2 r^2 + \beta^2 r^2 + 2\alpha \beta (\vec{u} \cdot \vec{v})$.
$1 = \alpha^2 + \beta^2 + 2\alpha \beta \frac{1}{\sqrt{2}} = \alpha^2 + \beta^2 + \sqrt{2} \alpha \beta$.
Substituting $\alpha = -\frac{\beta}{\sqrt{2}}$:
$1 = \frac{\beta^2}{2} + \beta^2 + \sqrt{2} (-\frac{\beta}{\sqrt{2}}) \beta = \frac{3\beta^2}{2} - \beta^2 = \frac{\beta^2}{2} \implies \beta^2 = 2$.
Then $\alpha^2 = \frac{2}{2} = 1$,so $\alpha = -1$ (since $\alpha = -\frac{\beta}{\sqrt{2}}$ and $\beta = \sqrt{2}$).
The roots are $\alpha = -1$ and $\beta^2 = 2$.
The quadratic equation is $(x - (-1))(x - 2) = (x+1)(x-2) = x^2 - x - 2 = 0$.

(D) Let the origin be at the circumcentre $P$. Then the position vectors of $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ such that $|\vec{a}| = |\vec{b}| = |\vec{c}| = R$,where $R$ is the circumradius.
The position vector of the orthocentre $Q$ is given by $\vec{q} = \vec{a} + \vec{b} + \vec{c}$.
Since $P$ is the origin,the position vector of $P$ is $\vec{p} = \vec{0}$.
We need to find $\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}$.
This is equal to $(\vec{a} - \vec{p}) + (\vec{b} - \vec{p}) + (\vec{c} - \vec{p}) = \vec{a} + \vec{b} + \vec{c} - 3\vec{p}$.
Since $\vec{p} = \vec{0}$,this simplifies to $\vec{a} + \vec{b} + \vec{c}$.
Since $\vec{q} = \vec{a} + \vec{b} + \vec{c}$ and $\vec{p} = \vec{0}$,we have $\vec{q} - \vec{p} = \vec{a} + \vec{b} + \vec{c}$.
Thus,$\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = \overrightarrow{PQ}$.

(B) Given $(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = \vec{0}$.
Since $\vec{c} \times \vec{c} = \vec{0}$,this simplifies to $(\vec{a} + \vec{b}) \times \vec{c} = \vec{0}$.
This implies $\vec{c}$ is parallel to $(\vec{a} + \vec{b})$. Let $\vec{c} = \alpha(\vec{a} + \vec{b})$ for some scalar $\alpha$.
$\vec{a} + \vec{b} = (\lambda + 3)\hat{i} + 0\hat{j} + 1\hat{k}$.
So,$\vec{c} = \alpha(\lambda + 3)\hat{i} + \alpha\hat{k}$.
Given $\vec{b} \cdot \vec{c} = -20 \Rightarrow (3\hat{i} - \hat{j} + 2\hat{k}) \cdot (\alpha(\lambda + 3)\hat{i} + \alpha\hat{k}) = -20$.
$3\alpha(\lambda + 3) + 2\alpha = -20 \Rightarrow \alpha(3\lambda + 11) = -20$.
Given $\vec{a} \cdot \vec{c} = -17 \Rightarrow (\lambda\hat{i} + \hat{j} - \hat{k}) \cdot (\alpha(\lambda + 3)\hat{i} + \alpha\hat{k}) = -17$.
$\alpha\lambda(\lambda + 3) - \alpha = -17 \Rightarrow \alpha(\lambda^2 + 3\lambda - 1) = -17$.
Dividing the two equations: $\frac{3\lambda + 11}{\lambda^2 + 3\lambda - 1} = \frac{20}{17}$.
$17(3\lambda + 11) = 20(\lambda^2 + 3\lambda - 1) \Rightarrow 51\lambda + 187 = 20\lambda^2 + 60\lambda - 20$.
$20\lambda^2 + 9\lambda - 207 = 0$. Solving for $\lambda \in \mathbb{Z}$,we get $\lambda = 3$.
Substituting $\lambda = 3$ into $\alpha(3(3) + 11) = -20 \Rightarrow 20\alpha = -20 \Rightarrow \alpha = -1$.
Thus,$\vec{c} = -1(6\hat{i} + \hat{k}) = -6\hat{i} - \hat{k}$.
We need to find $|\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})|^2$.
Let $\vec{v} = \vec{c} \times (3\hat{i} + \hat{j} + \hat{k}) = (-6\hat{i} - \hat{k}) \times (3\hat{i} + \hat{j} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -6 & 0 & -1 \\ 3 & 1 & 1 \end{vmatrix} = \hat{i}(0 - (-1)) - \hat{j}(-6 - (-3)) + \hat{k}(-6 - 0) = \hat{i} + 3\hat{j} - 6\hat{k}$.
$|\vec{v}|^2 = 1^2 + 3^2 + (-6)^2 = 1 + 9 + 36 = 46$.

(A) Given $\vec{d} \times \vec{b} = \vec{c} \times \vec{b}$.
This implies $(\vec{d} - \vec{c}) \times \vec{b} = 0$.
Therefore,$\vec{d} - \vec{c} = \lambda \vec{b}$,or $\vec{d} = \vec{c} + \lambda \vec{b}$ for some scalar $\lambda$.
Given $\vec{d} \cdot \vec{a} = 24$,substitute $\vec{d} = \vec{c} + \lambda \vec{b}$ into this equation:
$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 24 \Rightarrow \vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 24$.
Calculate the dot products:
$\vec{a} \cdot \vec{c} = (1)(2) + (4)(-1) + (2)(4) = 2 - 4 + 8 = 6$.
$\vec{a} \cdot \vec{b} = (1)(3) + (4)(-2) + (2)(7) = 3 - 8 + 14 = 9$.
Substitute these values: $6 + \lambda(9) = 24 \Rightarrow 9\lambda = 18 \Rightarrow \lambda = 2$.
Now,find $\vec{d} = \vec{c} + 2\vec{b} = (2\hat{i} - \hat{j} + 4\hat{k}) + 2(3\hat{i} - 2\hat{j} + 7\hat{k}) = (2+6)\hat{i} + (-1-4)\hat{j} + (4+14)\hat{k} = 8\hat{i} - 5\hat{j} + 18\hat{k}$.
Finally,calculate $|\vec{d}|^2 = 8^2 + (-5)^2 + 18^2 = 64 + 25 + 324 = 413$.

(A) Given that $\overline{AB} + \overline{BC} + \overline{CA} = \vec{0}$,we have $\overline{BC} = -\overline{AB} - \overline{CA}$.
$\overline{BC} = -(-2\hat{i} + \hat{j} + 3\hat{k}) - (4\hat{i} + 3\hat{j} + \delta\hat{k}) = -2\hat{i} - 4\hat{j} - (3 + \delta)\hat{k}$.
Since $\overline{CB} = -\overline{BC}$,we get $\overline{CB} = 2\hat{i} + 4\hat{j} + (3 + \delta)\hat{k}$.
Comparing with $\overline{CB} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$,we have $\alpha = 2$,$\beta = 4$,and $\gamma = 3 + \delta$.
The area of triangle $ABC$ is given by $\frac{1}{2} |\overline{AC} \times \overline{AB}| = 5\sqrt{6}$.
$\overline{AC} = -\overline{CA} = -4\hat{i} - 3\hat{j} - \delta\hat{k}$.
$\overline{AC} \times \overline{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & -3 & -\delta \\ -2 & 1 & 3 \end{vmatrix} = \hat{i}(-9 + \delta) - \hat{j}(-12 - 2\delta) + \hat{k}(-4 - 6) = (\delta - 9)\hat{i} + (2\delta + 12)\hat{j} - 10\hat{k}$.
$|\overline{AC} \times \overline{AB}|^2 = (\delta - 9)^2 + (2\delta + 12)^2 + (-10)^2 = (2\sqrt{5\sqrt{6}})^2 = 600$.
$(\delta^2 - 18\delta + 81) + (4\delta^2 + 48\delta + 144) + 100 = 600 \Rightarrow 5\delta^2 + 30\delta - 275 = 0 \Rightarrow \delta^2 + 6\delta - 55 = 0$.
$(\delta + 11)(\delta - 5) = 0$. Since $\delta > 0$,we have $\delta = 5$.
Then $\gamma = 3 + 5 = 8$,so $\overline{CB} = 2\hat{i} + 4\hat{j} + 8\hat{k}$ and $\overline{CA} = 4\hat{i} + 3\hat{j} + 5\hat{k}$.
$\overline{CB} \cdot \overline{CA} = (2)(4) + (4)(3) + (8)(5) = 8 + 12 + 40 = 60$.

(A) Let $\vec{a} = \alpha \hat{i}-2 \hat{j}+2 \hat{k}$ and $\vec{b} = \alpha \hat{i}+2 \alpha \hat{j}-2 \hat{k}$.
For the angle $\theta$ between the vectors to be acute,the dot product $\vec{a} \cdot \vec{b}$ must be greater than $0$.
$\vec{a} \cdot \vec{b} = (\alpha)(\alpha) + (-2)(2 \alpha) + (2)(-2) > 0$
$\alpha^2 - 4 \alpha - 4 > 0$
To solve $\alpha^2 - 4 \alpha - 4 = 0$,we use the quadratic formula $\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$\alpha = \frac{4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{4 \pm \sqrt{32}}{2} = 2 \pm 2 \sqrt{2}$
Since $2 \sqrt{2} \approx 2.828$,the roots are $\alpha_1 \approx 4.828$ and $\alpha_2 \approx -0.828$.
The inequality $\alpha^2 - 4 \alpha - 4 > 0$ holds for $\alpha > 2 + 2 \sqrt{2}$ or $\alpha < 2 - 2 \sqrt{2}$.
Since we are looking for the least positive integral value of $\alpha$,we consider $\alpha > 4.828$.
The smallest integer greater than $4.828$ is $5$.

(D) The position vectors of the vertices are $A(2, -3, 3)$,$B(2, 2, 3)$,and $C(-1, 1, 3)$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(2-2)^2 + (2 - (-3))^2 + (3-3)^2} = \sqrt{0^2 + 5^2 + 0^2} = 5$.
$AC = \sqrt{(-1-2)^2 + (1 - (-3))^2 + (3-3)^2} = \sqrt{(-3)^2 + 4^2 + 0^2} = \sqrt{9 + 16} = 5$.
Since $AB = AC = 5$,the triangle $ABC$ is an isosceles triangle.
In an isosceles triangle,the angle bisector $AD$ of the vertex angle $\angle BAC$ is also the median to the base $BC$.
Therefore,$D$ is the midpoint of $BC$.
$D = \left( \frac{2 + (-1)}{2}, \frac{2 + 1}{2}, \frac{3 + 3}{2} \right) = \left( \frac{1}{2}, \frac{3}{2}, 3 \right)$.
The length $l$ of the angle bisector $AD$ is the distance between $A(2, -3, 3)$ and $D\left(\frac{1}{2}, \frac{3}{2}, 3\right)$:
$l = \sqrt{\left(2 - \frac{1}{2}\right)^2 + \left(-3 - \frac{3}{2}\right)^2 + (3 - 3)^2}$
$l = \sqrt{\left(\frac{3}{2}\right)^2 + \left(-\frac{9}{2}\right)^2 + 0^2} = \sqrt{\frac{9}{4} + \frac{81}{4}} = \sqrt{\frac{90}{4}} = \sqrt{\frac{45}{2}}$.
Thus,$l^2 = \frac{45}{2}$.
Therefore,$2 l^2 = 2 \times \frac{45}{2} = 45$.

(D) The vertices are $P(3, 2, 3)$,$Q(4, 6, 2)$,and $R(7, 3, 2)$.
To find $\angle QPR$,we need the direction ratios of vectors $\vec{PQ}$ and $\vec{PR}$.
$\vec{PQ} = (4-3, 6-2, 2-3) = (1, 4, -1)$.
$\vec{PR} = (7-3, 3-2, 2-3) = (4, 1, -1)$.
Let $\theta = \angle QPR$. The cosine of the angle between two vectors $\vec{a}$ and $\vec{b}$ is given by $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
$\vec{PQ} \cdot \vec{PR} = (1)(4) + (4)(1) + (-1)(-1) = 4 + 4 + 1 = 9$.
$|\vec{PQ}| = \sqrt{1^2 + 4^2 + (-1)^2} = \sqrt{1 + 16 + 1} = \sqrt{18}$.
$|\vec{PR}| = \sqrt{4^2 + 1^2 + (-1)^2} = \sqrt{16 + 1 + 1} = \sqrt{18}$.
Therefore,$\cos \theta = \frac{9}{\sqrt{18} \cdot \sqrt{18}} = \frac{9}{18} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \frac{\pi}{3}$.

(B) Given unit vector $\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$.
Let $\overrightarrow{p}_1=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}$,$\overrightarrow{p}_2=\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$,and $\overrightarrow{p}_3=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$.
Since $\hat{u} \cdot \overrightarrow{p}_1 = |\hat{u}| |\overrightarrow{p}_1| \cos(\frac{\pi}{2}) = 0$,we have $\frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0 \Rightarrow x+z=0$ $(i)$.
Since $\hat{u} \cdot \overrightarrow{p}_2 = |\hat{u}| |\overrightarrow{p}_2| \cos(\frac{\pi}{3}) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}$,we have $\frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2} \Rightarrow y+z = \frac{1}{\sqrt{2}}$ $(ii)$.
Since $\hat{u} \cdot \overrightarrow{p}_3 = |\hat{u}| |\overrightarrow{p}_3| \cos(\frac{2\pi}{3}) = 1 \cdot 1 \cdot (-\frac{1}{2}) = -\frac{1}{2}$,we have $\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = -\frac{1}{2} \Rightarrow x+y = -\frac{1}{\sqrt{2}}$ $(iii)$.
Adding $(i), (ii), (iii)$,we get $2(x+y+z) = 0 \Rightarrow x+y+z = 0$.
Subtracting $(ii)$ from this,$x = -\frac{1}{\sqrt{2}}$. Subtracting $(iii)$ from this,$z = \frac{1}{\sqrt{2}}$. Subtracting $(i)$ from this,$y = 0$.
Thus $\hat{u} = -\frac{1}{\sqrt{2}} \hat{i} + 0 \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$.
Then $\hat{u}-\overrightarrow{v} = (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) \hat{i} + (0 - \frac{1}{\sqrt{2}}) \hat{j} + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) \hat{k} = -\frac{2}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} = -\sqrt{2} \hat{i} - \frac{1}{\sqrt{2}} \hat{j}$.
$|\hat{u}-\overrightarrow{v}|^2 = (-\sqrt{2})^2 + (-\frac{1}{\sqrt{2}})^2 = 2 + \frac{1}{2} = \frac{5}{2}$.

(A) Given $\vec{a} = \hat{i} + \alpha \hat{j} + \beta \hat{k}$,so $|\vec{a}|^2 = 1 + \alpha^2 + \beta^2$.
Given $|\vec{b}|^2 = 6$,so $|\vec{b}| = \sqrt{6}$.
The angle between $\vec{a}$ and $\vec{b}$ is $\theta = \frac{\pi}{4}$.
Using the dot product formula: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 3\sqrt{2}$.
Substituting the values: $|\vec{a}| \cdot \sqrt{6} \cdot \cos(\frac{\pi}{4}) = 3\sqrt{2}$.
$|\vec{a}| \cdot \sqrt{6} \cdot \frac{1}{\sqrt{2}} = 3\sqrt{2} \implies |\vec{a}| \cdot \sqrt{3} = 3\sqrt{2} \implies |\vec{a}| = \sqrt{6}$.
Thus,$|\vec{a}|^2 = 6$,which implies $1 + \alpha^2 + \beta^2 = 6$,so $\alpha^2 + \beta^2 = 5$.
Now,calculate $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$.
$|\vec{a} \times \vec{b}|^2 = (6)(6) \sin^2(\frac{\pi}{4}) = 36 \cdot (\frac{1}{\sqrt{2}})^2 = 36 \cdot \frac{1}{2} = 18$.
Finally,$(\alpha^2 + \beta^2) |\vec{a} \times \vec{b}|^2 = (5)(18) = 90$.

(D) Given $\vec{p} \times \vec{b} = \vec{c} \times \vec{b}$,we can write $\vec{p} \times \vec{b} - \vec{c} \times \vec{b} = \vec{0}$.
This implies $(\vec{p} - \vec{c}) \times \vec{b} = \vec{0}$.
Therefore,$\vec{p} - \vec{c} = \lambda \vec{b}$,which means $\vec{p} = \vec{c} + \lambda \vec{b}$ for some scalar $\lambda$.
Given $\vec{p} \cdot \vec{a} = 0$,we substitute $\vec{p}$:
$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 0 \Rightarrow \vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$.
Calculating the dot products:
$\vec{c} \cdot \vec{a} = (1)(3) + (-3)(1) + (4)(-2) = 3 - 3 - 8 = -8$.
$\vec{b} \cdot \vec{a} = (4)(3) + (1)(1) + (7)(-2) = 12 + 1 - 14 = -1$.
Substituting these values: $-8 + \lambda(-1) = 0 \Rightarrow \lambda = -8$.
Thus,$\vec{p} = \vec{c} - 8 \vec{b} = (\hat{i} - 3 \hat{j} + 4 \hat{k}) - 8(4 \hat{i} + \hat{j} + 7 \hat{k}) = (1-32) \hat{i} + (-3-8) \hat{j} + (4-56) \hat{k} = -31 \hat{i} - 11 \hat{j} - 52 \hat{k}$.
Finally,$\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k}) = (-31)(1) + (-11)(-1) + (-52)(-1) = -31 + 11 + 52 = 32$.

(A) Given vertices are $A(1,3,2)$,$B(-2,8,0)$,and $C(3,6,7)$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(-2-1)^2 + (8-3)^2 + (0-2)^2} = \sqrt{(-3)^2 + 5^2 + (-2)^2} = \sqrt{9 + 25 + 4} = \sqrt{38}$.
$AC = \sqrt{(3-1)^2 + (6-3)^2 + (7-2)^2} = \sqrt{2^2 + 3^2 + 5^2} = \sqrt{4 + 9 + 25} = \sqrt{38}$.
Since $AB = AC$,the triangle $ABC$ is isosceles,and the angle bisector $AD$ of $\angle BAC$ is also the median to $BC$. Thus,$D$ is the midpoint of $BC$.
$D = \left( \frac{-2+3}{2}, \frac{8+6}{2}, \frac{0+7}{2} \right) = \left( \frac{1}{2}, 7, \frac{7}{2} \right)$.
Now,find the vector $\overrightarrow{AD}$:
$\overrightarrow{AD} = \left( \frac{1}{2}-1 \right) \hat{i} + (7-3) \hat{j} + \left( \frac{7}{2}-2 \right) \hat{k} = -\frac{1}{2} \hat{i} + 4 \hat{j} + \frac{3}{2} \hat{k}$.
Find the vector $\overrightarrow{AC}$:
$\overrightarrow{AC} = (3-1) \hat{i} + (6-3) \hat{j} + (7-2) \hat{k} = 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$.
The length of the projection of $\overrightarrow{AD}$ on $\overrightarrow{AC}$ is given by $\left| \frac{\overrightarrow{AD} \cdot \overrightarrow{AC}}{|\overrightarrow{AC}|} \right|$.
$\overrightarrow{AD} \cdot \overrightarrow{AC} = \left( -\frac{1}{2} \right)(2) + (4)(3) + \left( \frac{3}{2} \right)(5) = -1 + 12 + 7.5 = 18.5 = \frac{37}{2}$.
$|\overrightarrow{AC}| = \sqrt{2^2 + 3^2 + 5^2} = \sqrt{38}$.
Projection length $= \left| \frac{37/2}{\sqrt{38}} \right| = \frac{37}{2 \sqrt{38}}$.

(A) Given $\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$,we have $(\vec{b} - \vec{c}) \times \vec{a} = 0$.
This implies $\vec{b} - \vec{c} = \lambda \vec{a}$ for some scalar $\lambda$.
Substituting the vectors: $(-\hat{i} - 8\hat{j} + 2\hat{k}) - (4\hat{i} + c_2\hat{j} + c_3\hat{k}) = \lambda(\hat{i} + \hat{j} + \hat{k})$.
Comparing components:
$-1 - 4 = \lambda \implies \lambda = -5$.
$-8 - c_2 = \lambda \implies -8 - c_2 = -5 \implies c_2 = -3$.
$2 - c_3 = \lambda \implies 2 - c_3 = -5 \implies c_3 = 7$.
Thus,$\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$.
Let $\vec{d} = 3\hat{i} + 4\hat{j} + \hat{k}$. Then $\cos \theta = \frac{\vec{c} \cdot \vec{d}}{|\vec{c}| |\vec{d}|} = \frac{(4)(3) + (-3)(4) + (7)(1)}{\sqrt{16+9+49} \sqrt{9+16+1}} = \frac{12 - 12 + 7}{\sqrt{74} \sqrt{26}} = \frac{7}{\sqrt{1924}} = \frac{7}{2\sqrt{481}}$.
$\cos^2 \theta = \frac{49}{4 \times 481} = \frac{49}{1924}$.
$\tan^2 \theta = \sec^2 \theta - 1 = \frac{1}{\cos^2 \theta} - 1 = \frac{1924}{49} - 1 = \frac{1875}{49} \approx 38.265$.
The greatest integer less than or equal to $\tan^2 \theta$ is $\lfloor 38.265 \rfloor = 38$.

(D) Let $\overrightarrow{C} = C_1 \hat{i} + C_2 \hat{j} + C_3 \hat{k}$ be a unit vector,so $C_1^2 + C_2^2 + C_3^2 = 1$.
Given $\overrightarrow{C} \cdot (2 \hat{i} + 2 \hat{j} - \hat{k}) = |\overrightarrow{C}| |2 \hat{i} + 2 \hat{j} - \hat{k}| \cos 60^{\circ} = 1 \cdot 3 \cdot \frac{1}{2} = \frac{3}{2}$.
So,$2C_1 + 2C_2 - C_3 = \frac{3}{2}$.
Also,$\overrightarrow{C} \cdot (\hat{i} - \hat{k}) = |\overrightarrow{C}| |\hat{i} - \hat{k}| \cos 45^{\circ} = 1 \cdot \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1$.
So,$C_1 - C_3 = 1$,which implies $C_3 = C_1 - 1$.
Substituting $C_3$ into the first equation: $2C_1 + 2C_2 - (C_1 - 1) = \frac{3}{2} \implies C_1 + 2C_2 = \frac{1}{2} \implies C_2 = \frac{1}{4} - \frac{C_1}{2}$.
Using $C_1^2 + C_2^2 + C_3^2 = 1$,we get $C_1^2 + (\frac{1}{4} - \frac{C_1}{2})^2 + (C_1 - 1)^2 = 1$.
Solving this quadratic equation yields $C_1 = \frac{1}{2} + \frac{\sqrt{2}}{3}$,$C_2 = -\frac{1}{3\sqrt{2}}$,and $C_3 = \frac{\sqrt{2}}{3} - \frac{1}{2}$.
Thus,$\overrightarrow{C} = (\frac{1}{2} + \frac{\sqrt{2}}{3}) \hat{i} - \frac{1}{3\sqrt{2}} \hat{j} + (\frac{\sqrt{2}}{3} - \frac{1}{2}) \hat{k}$.
Adding the given vector $(-\frac{1}{2} \hat{i} + \frac{1}{3\sqrt{2}} \hat{j} - \frac{\sqrt{2}}{3} \hat{k})$ to $\overrightarrow{C}$ results in $\frac{\sqrt{2}}{3} \hat{i} + 0 \hat{j} - \frac{1}{2} \hat{k} = \frac{\sqrt{2}}{3} \hat{i} - \frac{1}{2} \hat{k}$.

(A) Given that $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are perpendicular,their dot product is zero: $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0$.
This simplifies to $|\vec{a}|^2 - |\vec{b}|^2 = 0$,which means $|\vec{a}|^2 = |\vec{b}|^2$.
Calculating the magnitudes: $|\vec{a}|^2 = 1^2 + \lambda^2 + (-3)^2 = 10 + \lambda^2$ and $|\vec{b}|^2 = 3^2 + (-1)^2 + 2^2 = 9 + 1 + 4 = 14$.
Equating them: $10 + \lambda^2 = 14 \implies \lambda^2 = 4$. Since $\lambda > 0$,we have $\lambda = 2$.
Now,$\vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$.
The dot product $\vec{a} \cdot \vec{b} = (1)(3) + (2)(-1) + (-3)(2) = 3 - 2 - 6 = -5$.
The angle $\theta$ is given by $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
Since $|\vec{a}|^2 = 14$ and $|\vec{b}|^2 = 14$,we have $|\vec{a}| = \sqrt{14}$ and $|\vec{b}| = \sqrt{14}$.
Thus,$\cos \theta = \frac{-5}{\sqrt{14} \cdot \sqrt{14}} = \frac{-5}{14}$.
Therefore,$14 \cos \theta = -5$,and $(14 \cos \theta)^2 = (-5)^2 = 25$.

(D) Given $(\vec{a}+2 \vec{b}) \times \vec{c}=3(\vec{c} \times \vec{a})$.
Since $\vec{c} \times \vec{a} = -(\vec{a} \times \vec{c})$,we have $(\vec{a}+2 \vec{b}) \times \vec{c} = -3(\vec{a} \times \vec{c})$.
$(\vec{a}+2 \vec{b}) \times \vec{c} + 3(\vec{a} \times \vec{c}) = 0$.
$(\vec{a} + 2\vec{b} + 3\vec{a}) \times \vec{c} = 0$.
$(4\vec{a} + 2\vec{b}) \times \vec{c} = 0$.
This implies $\vec{c}$ is parallel to $(4\vec{a} + 2\vec{b})$.
Let $\vec{c} = \lambda(4\vec{a} + 2\vec{b})$.
$4\vec{a} + 2\vec{b} = 4(\hat{i}-3\hat{j}+7\hat{k}) + 2(2\hat{i}-\hat{j}+\hat{k}) = (4+4)\hat{i} + (-12-2)\hat{j} + (28+2)\hat{k} = 8\hat{i} - 14\hat{j} + 30\hat{k}$.
So,$\vec{c} = \lambda(8\hat{i} - 14\hat{j} + 30\hat{k})$.
Given $\vec{a} \cdot \vec{c} = 130$.
$(\hat{i}-3\hat{j}+7\hat{k}) \cdot \lambda(8\hat{i}-14\hat{j}+30\hat{k}) = 130$.
$\lambda(8 + 42 + 210) = 130$.
$260\lambda = 130 \implies \lambda = \frac{1}{2}$.
Thus,$\vec{c} = 4\hat{i} - 7\hat{j} + 15\hat{k}$.
Finally,$\vec{b} \cdot \vec{c} = (2\hat{i}-\hat{j}+\hat{k}) \cdot (4\hat{i}-7\hat{j}+15\hat{k}) = 8 + 7 + 15 = 30$.

(B) Given $\vec{a} = 2\hat{i} + 5\hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} - 2\hat{j} + 2\hat{k}$.
Let $\vec{v} = \vec{a} + \vec{b} + \hat{i} = (2+2+1)\hat{i} + (5-2)\hat{j} + (-1+2)\hat{k} = 5\hat{i} + 3\hat{j} + \hat{k}$.
Let $\vec{p} = \vec{c} + \hat{i}$.
The given equation is $\vec{p} \times \vec{v} = \vec{a} \times \vec{p}$.
This implies $\vec{p} \times \vec{v} + \vec{p} \times \vec{a} = \vec{0}$,so $\vec{p} \times (\vec{v} + \vec{a}) = \vec{0}$.
Thus,$\vec{p} = \lambda(\vec{v} + \vec{a})$ for some scalar $\lambda$.
$\vec{v} + \vec{a} = (5\hat{i} + 3\hat{j} + \hat{k}) + (2\hat{i} + 5\hat{j} - \hat{k}) = 7\hat{i} + 8\hat{j}$.
So,$\vec{c} + \hat{i} = \lambda(7\hat{i} + 8\hat{j}) \Rightarrow \vec{c} = 7\lambda\hat{i} + 8\lambda\hat{j} - \hat{i} = (7\lambda - 1)\hat{i} + 8\lambda\hat{j}$.
Given $\vec{a} \cdot \vec{c} = -29$,we have $(2\hat{i} + 5\hat{j} - \hat{k}) \cdot ((7\lambda - 1)\hat{i} + 8\lambda\hat{j}) = -29$.
$2(7\lambda - 1) + 5(8\lambda) = -29 \Rightarrow 14\lambda - 2 + 40\lambda = -29 \Rightarrow 54\lambda = -27 \Rightarrow \lambda = -\frac{1}{2}$.
Now,$\vec{c} = (7(-\frac{1}{2}) - 1)\hat{i} + 8(-\frac{1}{2})\hat{j} = -\frac{9}{2}\hat{i} - 4\hat{j}$.
We need to find $\vec{c} \cdot (-2\hat{i} + \hat{j} + \hat{k}) = (-\frac{9}{2}\hat{i} - 4\hat{j}) \cdot (-2\hat{i} + \hat{j} + \hat{k}) = (-\frac{9}{2})(-2) + (-4)(1) + (0)(1) = 9 - 4 = 5$.

(C) Given $\overrightarrow{a} = \overrightarrow{b} \times \overrightarrow{c}$,it implies $\overrightarrow{a} \perp \overrightarrow{b}$ and $\overrightarrow{a} \perp \overrightarrow{c}$. Thus,$\overrightarrow{a} \cdot \overrightarrow{c} = 0$.
We have $|\overrightarrow{c} - \overrightarrow{a}|^2 = |\overrightarrow{c}|^2 + |\overrightarrow{a}|^2 - 2(\overrightarrow{a} \cdot \overrightarrow{c}) = |\overrightarrow{c}|^2 + 2^2 - 0 = |\overrightarrow{c}|^2 + 4$.
From $|\overrightarrow{a}| = |\overrightarrow{b} \times \overrightarrow{c}|$,we get $2 = |\overrightarrow{b}| |\overrightarrow{c}| \sin \alpha = 3 |\overrightarrow{c}| \sin \alpha$.
Therefore,$|\overrightarrow{c}| = \frac{2}{3 \sin \alpha} = \frac{2}{3} \csc \alpha$.
Since $\alpha \in [0, \frac{\pi}{3}]$,the range of $\csc \alpha$ is $[\frac{2}{\sqrt{3}}, \infty)$.
The minimum value of $|\overrightarrow{c}|$ occurs at $\alpha = \frac{\pi}{3}$,where $|\overrightarrow{c}| = \frac{2}{3} \times \frac{2}{\sqrt{3}} = \frac{4}{3\sqrt{3}}$.
Thus,$|\overrightarrow{c}|^2 = \frac{16}{27}$.
Substituting this into the expression,$27|\overrightarrow{c} - \overrightarrow{a}|^2 = 27(|\overrightarrow{c}|^2 + 4) = 27(\frac{16}{27} + 4) = 16 + 108 = 124$.

(C) Two vectors $\vec{a}$ and $\vec{b}$ are inclined at an obtuse angle if their dot product $\vec{a} \cdot \vec{b} < 0$.
Given $\vec{a} = \alpha t \hat{i} + 6 \hat{j} - 3 \hat{k}$ and $\vec{b} = t \hat{i} - 2 \hat{j} - 2 \alpha t \hat{k}$.
Calculating the dot product: $\vec{a} \cdot \vec{b} = (\alpha t)(t) + (6)(-2) + (-3)(-2 \alpha t) = \alpha t^2 - 12 + 6 \alpha t$.
We require $\alpha t^2 + 6 \alpha t - 12 < 0$ for all $t \in R$.
For a quadratic expression $f(t) = At^2 + Bt + C$ to be negative for all $t$,we must have $A < 0$ and the discriminant $D = B^2 - 4AC < 0$.
Here $A = \alpha$,$B = 6 \alpha$,and $C = -12$.
Condition $1$: $\alpha < 0$.
Condition $2$: $D = (6 \alpha)^2 - 4(\alpha)(-12) = 36 \alpha^2 + 48 \alpha < 0$.
$12 \alpha (3 \alpha + 4) < 0$,which implies $-\frac{4}{3} < \alpha < 0$.
If $\alpha = 0$,the expression becomes $-12 < 0$,which is true for all $t \in R$.
Combining these,the set of all $\alpha$ is $(-\frac{4}{3}, 0]$.

(D) Given $\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}$,$\vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$,and $\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}$.
From $\vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a}$,we have $(\vec{r}-(\vec{b}+\vec{c})) \times \vec{a} = 0$.
This implies $\vec{r}-(\vec{b}+\vec{c}) = \lambda \vec{a}$ for some scalar $\lambda$,so $\vec{r} = \lambda \vec{a} + \vec{b} + \vec{c}$.
Given $\vec{r} \cdot (\vec{b}-\vec{c}) = 0$,substitute $\vec{r}$:
$(\lambda \vec{a} + \vec{b} + \vec{c}) \cdot (\vec{b}-\vec{c}) = 0$.
$\lambda (\vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}) + (\vec{b} + \vec{c}) \cdot (\vec{b} - \vec{c}) = 0$.
$\lambda (\vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}) + |\vec{b}|^2 - |\vec{c}|^2 = 0$.
Calculate dot products: $\vec{a} \cdot \vec{b} = (9)(3) + (-13)(7) + (25)(-13) = 27 - 91 - 325 = -389$.
$\vec{a} \cdot \vec{c} = (9)(17) + (-13)(-2) + (25)(1) = 153 + 26 + 25 = 204$.
$|\vec{b}|^2 = 3^2 + 7^2 + (-13)^2 = 9 + 49 + 169 = 227$.
$|\vec{c}|^2 = 17^2 + (-2)^2 + 1^2 = 289 + 4 + 1 = 294$.
$\lambda (-389 - 204) + 227 - 294 = 0 \Rightarrow -593 \lambda - 67 = 0 \Rightarrow \lambda = -\frac{67}{593}$.
Thus,$\vec{r} = -\frac{67}{593} \vec{a} + \vec{b} + \vec{c}$.
$593 \vec{r} + 67 \vec{a} = 593(\vec{b} + \vec{c})$.
$\frac{|593 \vec{r} + 67 \vec{a}|^2}{(593)^2} = |\vec{b} + \vec{c}|^2$.
$\vec{b} + \vec{c} = (3+17)\hat{i} + (7-2)\hat{j} + (-13+1)\hat{k} = 20\hat{i} + 5\hat{j} - 12\hat{k}$.
$|\vec{b} + \vec{c}|^2 = 20^2 + 5^2 + (-12)^2 = 400 + 25 + 144 = 569$.

(B) Given $(\vec{a}+\vec{b}) \times \vec{c} = \vec{c} \times (-2 \vec{a}+3 \vec{b})$.
This can be rewritten as $(\vec{a}+\vec{b}) \times \vec{c} + (-2 \vec{a}+3 \vec{b}) \times \vec{c} = \vec{0}$.
$(\vec{a}+\vec{b}-2 \vec{a}+3 \vec{b}) \times \vec{c} = \vec{0}$.
$(4 \vec{b}-\vec{a}) \times \vec{c} = \vec{0}$.
This implies $\vec{c}$ is parallel to $(4 \vec{b}-\vec{a})$.
Let $\vec{c} = \lambda(4 \vec{b}-\vec{a})$.
$4 \vec{b}-\vec{a} = 4(11 \hat{i}-\hat{j}+\hat{k}) - (4 \hat{i}-\hat{j}+\hat{k}) = (44-4)\hat{i} + (-4+1)\hat{j} + (4-1)\hat{k} = 40 \hat{i}-3 \hat{j}+3 \hat{k}$.
So,$\vec{c} = \lambda(40 \hat{i}-3 \hat{j}+3 \hat{k})$.
Given $(2 \vec{a}+3 \vec{b}) \cdot \vec{c} = 1670$.
$2 \vec{a}+3 \vec{b} = 2(4 \hat{i}-\hat{j}+\hat{k}) + 3(11 \hat{i}-\hat{j}+\hat{k}) = (8+33)\hat{i} + (-2-3)\hat{j} + (2+3)\hat{k} = 41 \hat{i}-5 \hat{j}+5 \hat{k}$.
$(41 \hat{i}-5 \hat{j}+5 \hat{k}) \cdot \lambda(40 \hat{i}-3 \hat{j}+3 \hat{k}) = 1670$.
$\lambda(41 \times 40 + (-5) \times (-3) + 5 \times 3) = 1670$.
$\lambda(1640 + 15 + 15) = 1670$.
$1670 \lambda = 1670 \Rightarrow \lambda = 1$.
Thus,$\vec{c} = 40 \hat{i}-3 \hat{j}+3 \hat{k}$.
$|\vec{c}|^2 = 40^2 + (-3)^2 + 3^2 = 1600 + 9 + 9 = 1618$.