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

(C) Given that $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 12$.
Using the distributive property of the dot product:
$\vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a} = 12$.
Since the dot product is commutative,$\vec{x} \cdot \vec{a} = \vec{a} \cdot \vec{x}$,so they cancel out:
$|\vec{x}|^2 - |\vec{a}|^2 = 12$.
Given that $\vec{a}$ is a unit vector,$|\vec{a}| = 1$,so $|\vec{a}|^2 = 1$.
Substituting this into the equation:
$|\vec{x}|^2 - 1 = 12$.
$|\vec{x}|^2 = 13$.
Taking the square root on both sides,we get $|\vec{x}| = \sqrt{13}$.

(A) Given vectors are $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}.$
First,calculate the vector $\vec{a}+\lambda \vec{b}$:
$\vec{a}+\lambda \vec{b} = (2 \hat{i}+2 \hat{j}+3 \hat{k}) + \lambda(-\hat{i}+2 \hat{j}+\hat{k})$
$= (2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}.$
Since $(\vec{a}+\lambda \vec{b})$ is perpendicular to $\vec{c}$,their dot product must be zero:
$(\vec{a}+\lambda \vec{b}) \cdot \vec{c} = 0$
$[(2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}] \cdot (3 \hat{i} + \hat{j} + 0 \hat{k}) = 0$
Calculating the dot product:
$3(2-\lambda) + 1(2+2\lambda) + 0(3+\lambda) = 0$
$6 - 3\lambda + 2 + 2\lambda = 0$
$8 - \lambda = 0$
$\lambda = 8.$

To show that two vectors are perpendicular,their dot product must be equal to $0$.
Let $\vec{u} = |\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ and $\vec{v} = |\vec{a}| \vec{b}-|\vec{b}| \vec{a}$.
Calculate the dot product $\vec{u} \cdot \vec{v}$:
$\vec{u} \cdot \vec{v} = (|\vec{a}| \vec{b}+|\vec{b}| \vec{a}) \cdot (|\vec{a}| \vec{b}-|\vec{b}| \vec{a})$
Using the distributive property of the dot product:
$= |\vec{a}|^2 (\vec{b} \cdot \vec{b}) - |\vec{a}||\vec{b}| (\vec{b} \cdot \vec{a}) + |\vec{b}||\vec{a}| (\vec{a} \cdot \vec{b}) - |\vec{b}|^2 (\vec{a} \cdot \vec{a})$
Since $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ and $\vec{x} \cdot \vec{x} = |\vec{x}|^2$:
$= |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}||\vec{b}| (\vec{a} \cdot \vec{b}) + |\vec{a}||\vec{b}| (\vec{a} \cdot \vec{b}) - |\vec{b}|^2 |\vec{a}|^2$
$= |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}|^2 |\vec{b}|^2$
$= 0$
Since the dot product is $0$,the vectors $|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ and $|\vec{a}| \vec{b}-|\vec{b}| \vec{a}$ are perpendicular to each other.

(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1.$
We are given the equation $\vec{a}+\vec{b}+\vec{c}=\vec{0}.$
Squaring both sides,we get:
$|\vec{a}+\vec{b}+\vec{c}|^{2} = |\vec{0}|^{2}$
Expanding the dot product:
$(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) = 0$
$|\vec{a}|^{2} + |\vec{b}|^{2} + |\vec{c}|^{2} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
Substituting the magnitudes $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$:
$1^{2} + 1^{2} + 1^{2} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
$3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
Solving for the required expression:
$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3$
$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$

(N/A) Consider $\vec{a}=2 \hat{i}+4 \hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}+3 \hat{j}-6 \hat{k}.$
Then,the dot product is:
$\vec{a} \cdot \vec{b}=(2)(3)+(4)(3)+(3)(-6)=6+12-18=0.$
We now observe the magnitudes:
$|\vec{a}|=\sqrt{2^{2}+4^{2}+3^{2}}=\sqrt{4+16+9}=\sqrt{29} \neq 0.$
$|\vec{b}|=\sqrt{3^{2}+3^{2}+(-6)^{2}}=\sqrt{9+9+36}=\sqrt{54} \neq 0.$
Since $\vec{a} \cdot \vec{b}=0$ but both $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0},$ the converse of the statement is not necessarily true.

(A) The vertices of $\triangle ABC$ are $A(1,2,3), B(-1,0,0),$ and $C(0,1,2)$.
$\angle ABC$ is the angle between the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$.
$\overrightarrow{BA} = (1 - (-1))\hat{i} + (2 - 0)\hat{j} + (3 - 0)\hat{k} = 2\hat{i} + 2\hat{j} + 3\hat{k}$.
$\overrightarrow{BC} = (0 - (-1))\hat{i} + (1 - 0)\hat{j} + (2 - 0)\hat{k} = \hat{i} + \hat{j} + 2\hat{k}$.
$\overrightarrow{BA} \cdot \overrightarrow{BC} = (2)(1) + (2)(1) + (3)(2) = 2 + 2 + 6 = 10$.
$|\overrightarrow{BA}| = \sqrt{2^2 + 2^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$.
$|\overrightarrow{BC}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
Using the dot product formula: $\cos(\angle ABC) = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}| |\overrightarrow{BC}|} = \frac{10}{\sqrt{17} \times \sqrt{6}} = \frac{10}{\sqrt{102}}$.
Therefore,$\angle ABC = \cos^{-1}\left(\frac{10}{\sqrt{102}}\right)$.

Let the position vectors of points $A, B,$ and $C$ be $\overrightarrow{OA} = 2 \hat{i} - \hat{j} + \hat{k}$,$\overrightarrow{OB} = \hat{i} - 3 \hat{j} - 5 \hat{k}$,and $\overrightarrow{OC} = 3 \hat{i} - 4 \hat{j} - 4 \hat{k}$.
The vectors representing the sides of $\Delta ABC$ are:
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (1-2) \hat{i} + (-3+1) \hat{j} + (-5-1) \hat{k} = -\hat{i} - 2 \hat{j} - 6 \hat{k}$
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (3-1) \hat{i} + (-4+3) \hat{j} + (-4+5) \hat{k} = 2 \hat{i} - \hat{j} + \hat{k}$
$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (3-2) \hat{i} + (-4+1) \hat{j} + (-4-1) \hat{k} = \hat{i} - 3 \hat{j} - 5 \hat{k}$
Now,calculate the squares of the magnitudes of the sides:
$|\overrightarrow{AB}|^2 = (-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = 41$
$|\overrightarrow{BC}|^2 = (2)^2 + (-1)^2 + (1)^2 = 4 + 1 + 1 = 6$
$|\overrightarrow{AC}|^2 = (1)^2 + (-3)^2 + (-5)^2 = 1 + 9 + 25 = 35$
Since $|\overrightarrow{BC}|^2 + |\overrightarrow{AC}|^2 = 6 + 35 = 41 = |\overrightarrow{AB}|^2$,the sides satisfy the Pythagorean theorem.
Therefore,$\Delta ABC$ is a right-angled triangle.

(A) The given lines are in the form $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$.
Here,$\vec{b}_1 = \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $\vec{b}_2 = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$.
The angle $\theta$ between the two lines is given by $\cos \theta = \left| \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1| |\vec{b}_2|} \right|$.
First,calculate the dot product: $\vec{b}_1 \cdot \vec{b}_2 = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19$.
Next,calculate the magnitudes: $|\vec{b}_1| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$|\vec{b}_2| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
Substituting these values into the formula: $\cos \theta = \left| \frac{19}{3 \times 7} \right| = \frac{19}{21}$.
Therefore,$\theta = \cos^{-1}\left(\frac{19}{21}\right)$.

(A) The two lines are parallel because their direction vectors are identical,$\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.
We have $\vec{a}_{1} = \hat{i} + 2\hat{j} - 4\hat{k}$ and $\vec{a}_{2} = 3\hat{i} + 3\hat{j} - 5\hat{k}$.
The distance $d$ between two parallel lines is given by $d = \frac{|\vec{b} \times (\vec{a}_{2} - \vec{a}_{1})|}{|\vec{b}|}$.
First,calculate $\vec{a}_{2} - \vec{a}_{1} = (3-1)\hat{i} + (3-2)\hat{j} + (-5 - (-4))\hat{k} = 2\hat{i} + \hat{j} - \hat{k}$.
Next,calculate the cross product $\vec{b} \times (\vec{a}_{2} - \vec{a}_{1}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 2 & 1 & -1 \end{vmatrix} = \hat{i}(-3-6) - \hat{j}(-2-12) + \hat{k}(2-6) = -9\hat{i} + 14\hat{j} - 4\hat{k}$.
The magnitude is $|\vec{b} \times (\vec{a}_{2} - \vec{a}_{1})| = \sqrt{(-9)^2 + 14^2 + (-4)^2} = \sqrt{81 + 196 + 16} = \sqrt{293}$.
The magnitude of $\vec{b}$ is $|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
Thus,the distance is $d = \frac{\sqrt{293}}{7}$.

(A) Let $Q$ be the angle between the given lines.
The angle between the given lines is given by the formula,$\cos Q = \left| \frac{\vec{b}_{1} \cdot \vec{b}_{2}}{|\vec{b}_{1}| |\vec{b}_{2}|} \right|$.
The given lines are parallel to the vectors $\vec{b}_{1} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$ and $\vec{b}_{2} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ respectively.
Calculate the magnitudes of the vectors:
$|\vec{b}_{1}| = \sqrt{3^{2} + 2^{2} + 6^{2}} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
$|\vec{b}_{2}| = \sqrt{1^{2} + 2^{2} + 2^{2}} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
Calculate the dot product of the vectors:
$\vec{b}_{1} \cdot \vec{b}_{2} = (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) \cdot (\hat{i} + 2 \hat{j} + 2 \hat{k})$
$= (3 \times 1) + (2 \times 2) + (6 \times 2) = 3 + 4 + 12 = 19$.
Substitute the values into the formula:
$\cos Q = \frac{19}{7 \times 3} = \frac{19}{21}$.
Therefore,$Q = \cos ^{-1}\left(\frac{19}{21}\right)$.

(A) The given lines are parallel to the vectors $\vec{b}_{1}=\hat{i}-\hat{j}-2\hat{k}$ and $\vec{b}_{2}=3\hat{i}-5\hat{j}-4\hat{k}$ respectively.
First,calculate the magnitudes of the vectors:
$|\vec{b}_{1}|=\sqrt{(1)^{2}+(-1)^{2}+(-2)^{2}}=\sqrt{1+1+4}=\sqrt{6}$
$|\vec{b}_{2}|=\sqrt{(3)^{2}+(-5)^{2}+(-4)^{2}}=\sqrt{9+25+16}=\sqrt{50}=5\sqrt{2}$
Next,calculate the dot product $\vec{b}_{1} \cdot \vec{b}_{2}$:
$\vec{b}_{1} \cdot \vec{b}_{2} = (1)(3) + (-1)(-5) + (-2)(-4) = 3 + 5 + 8 = 16$
The angle $Q$ between the lines is given by:
$\cos Q = \frac{|\vec{b}_{1} \cdot \vec{b}_{2}|}{|\vec{b}_{1}| |\vec{b}_{2}|}$
Substituting the values:
$\cos Q = \frac{16}{\sqrt{6} \cdot 5\sqrt{2}} = \frac{16}{\sqrt{2} \cdot \sqrt{3} \cdot 5\sqrt{2}} = \frac{16}{5 \cdot 2 \cdot \sqrt{3}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}$
Therefore,$Q = \cos^{-1}\left(\frac{8}{5\sqrt{3}}\right)$.

(D) Let the direction ratios of the two lines be $l_1 = (a, b, c)$ and $l_2 = (b-c, c-a, a-b)$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Calculate the dot product of the direction ratios:
$a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ba + ca - cb = 0$.
Since the dot product is $0$,the numerator of the formula is $0$.
Therefore,$\cos \theta = 0$,which implies $\theta = 90^{\circ}$.

(N/A) Let $\vec{a} = l_{1}\hat{i} + m_{1}\hat{j} + n_{1}\hat{k}$,$\vec{b} = l_{2}\hat{i} + m_{2}\hat{j} + n_{2}\hat{k}$,and $\vec{c} = l_{3}\hat{i} + m_{3}\hat{j} + n_{3}\hat{k}$ be the unit vectors along the three mutually perpendicular lines.
Let $\vec{d} = (l_{1}+l_{2}+l_{3})\hat{i} + (m_{1}+m_{2}+m_{3})\hat{j} + (n_{1}+n_{2}+n_{3})\hat{k} = \vec{a} + \vec{b} + \vec{c}$.
Let $\alpha, \beta, \gamma$ be the angles between $\vec{d}$ and $\vec{a}, \vec{b}, \vec{c}$ respectively.
Then $\cos \alpha = \frac{\vec{a} \cdot \vec{d}}{|\vec{a}| |\vec{d}|} = \frac{\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c})}{|\vec{d}|} = \frac{\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}}{|\vec{d}|}$.
Since the lines are mutually perpendicular,$\vec{a} \cdot \vec{b} = 0$,$\vec{a} \cdot \vec{c} = 0$,and $\vec{a} \cdot \vec{a} = 1$.
Thus,$\cos \alpha = \frac{1}{|\vec{d}|}$.
Similarly,$\cos \beta = \frac{\vec{b} \cdot (\vec{a} + \vec{b} + \vec{c})}{|\vec{d}|} = \frac{1}{|\vec{d}|}$ and $\cos \gamma = \frac{\vec{c} \cdot (\vec{a} + \vec{b} + \vec{c})}{|\vec{d}|} = \frac{1}{|\vec{d}|}$.
Since $\cos \alpha = \cos \beta = \cos \gamma$,it follows that $\alpha = \beta = \gamma$. Hence,the line makes equal angles with the three mutually perpendicular lines.

(A) It is given that $|\vec{a}|=3$ and $|\vec{b}|=\frac{\sqrt{2}}{3}$.
We know that the magnitude of the cross product is given by $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
Since $\vec{a} \times \vec{b}$ is a unit vector,its magnitude must be $1$,i.e.,$|\vec{a} \times \vec{b}| = 1$.
Substituting the given values,we get:
$3 \times \frac{\sqrt{2}}{3} \times \sin \theta = 1$
$\sqrt{2} \sin \theta = 1$
$\sin \theta = \frac{1}{\sqrt{2}}$
Since $\sin \theta = \frac{1}{\sqrt{2}}$,the angle $\theta = \frac{\pi}{4}$.
Thus,$\vec{a} \times \vec{b}$ is a unit vector if the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$.

(D) Let $\theta$ be the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$.
First,calculate the vectors:
$\overrightarrow{AB} = (2\hat{i} + 5\hat{j}) - (\hat{i} + \hat{j} + \hat{k}) = \hat{i} + 4\hat{j} - \hat{k}$
$\overrightarrow{CD} = (\hat{i} - 6\hat{j} - \hat{k}) - (3\hat{i} + 2\hat{j} - 3\hat{k}) = -2\hat{i} - 8\hat{j} + 2\hat{k}$
Calculate the magnitudes:
$|\overrightarrow{AB}| = \sqrt{1^2 + 4^2 + (-1)^2} = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2}$
$|\overrightarrow{CD}| = \sqrt{(-2)^2 + (-8)^2 + 2^2} = \sqrt{4 + 64 + 4} = \sqrt{72} = 6\sqrt{2}$
Calculate the dot product:
$\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(-2) + (4)(-8) + (-1)(2) = -2 - 32 - 2 = -36$
Calculate the angle $\cos \theta$:
$\cos \theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{AB}| |\overrightarrow{CD}|} = \frac{-36}{(3\sqrt{2})(6\sqrt{2})} = \frac{-36}{18 \times 2} = \frac{-36}{36} = -1$
Since $\cos \theta = -1$,we have $\theta = \pi$.
Since the angle between the vectors is $\pi$,the vectors are collinear (specifically,they are in opposite directions). Alternatively,$\overrightarrow{CD} = -2(\hat{i} + 4\hat{j} - \hat{k}) = -2\overrightarrow{AB}$,which confirms they are collinear.

(A) Given that $\vec{a} \cdot (\vec{b}+\vec{c}) = 0$,$\vec{b} \cdot (\vec{c}+\vec{a}) = 0$,and $\vec{c} \cdot (\vec{a}+\vec{b}) = 0$.
We know that $|\vec{a}+\vec{b}+\vec{c}|^2 = (\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c})$.
Expanding this,we get:
$|\vec{a}+\vec{b}+\vec{c}|^2 = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.
From the given conditions:
$\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0$
$\vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0$
$\vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0$
Adding these three equations,we get $2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.
Thus,$|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2$.
Substituting the given magnitudes:
$|\vec{a}+\vec{b}+\vec{c}|^2 = 3^2 + 4^2 + 5^2 = 9 + 16 + 25 = 50$.
Therefore,$|\vec{a}+\vec{b}+\vec{c}| = \sqrt{50} = 5\sqrt{2}$.

(B) Given $\vec{a}+\vec{b}+\vec{c}=\vec{0}.$
Squaring both sides,we get $(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) = 0.$
Expanding the dot product,we have $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0.$
Substitute the given magnitudes $|\vec{a}|=1, |\vec{b}|=4, |\vec{c}|=2:$
$1^2 + 4^2 + 2^2 + 2\mu = 0.$
$1 + 16 + 4 + 2\mu = 0.$
$21 + 2\mu = 0.$
$2\mu = -21.$
$\mu = -\frac{21}{2}.$

Let $\vec{\beta}_{1} = \lambda \vec{\alpha}$,where $\lambda$ is a scalar. Then $\vec{\beta}_{1} = 3\lambda \hat{i} - \lambda \hat{j}$.
Now,$\vec{\beta}_{2} = \vec{\beta} - \vec{\beta}_{1} = (2 - 3\lambda) \hat{i} + (1 + \lambda) \hat{j} - 3\hat{k}$.
Since $\vec{\beta}_{2}$ is perpendicular to $\vec{\alpha}$,we have $\vec{\alpha} \cdot \vec{\beta}_{2} = 0$.
$3(2 - 3\lambda) - 1(1 + \lambda) = 0$
$6 - 9\lambda - 1 - \lambda = 0$
$5 - 10\lambda = 0 \implies \lambda = \frac{1}{2}$.
Thus,$\vec{\beta}_{1} = \frac{3}{2}\hat{i} - \frac{1}{2}\hat{j}$ and $\vec{\beta}_{2} = \frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}$.

(A) Let $\vec{a} = \hat{i}+\hat{j}+\hat{k}$.
Let $\vec{b} = (2 \hat{i}+4 \hat{j}-5 \hat{k}) + (\lambda \hat{i}+2 \hat{j}+3 \hat{k}) = (2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}$.
The unit vector along $\vec{b}$ is $\hat{b} = \frac{\vec{b}}{|\vec{b}|} = \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^{2}+6^{2}+(-2)^{2}}} = \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^{2}+4 \lambda+44}}$.
The scalar product $\vec{a} \cdot \hat{b} = 1$.
$\Rightarrow \frac{(2+\lambda)(1) + (6)(1) + (-2)(1)}{\sqrt{\lambda^{2}+4 \lambda+44}} = 1$.
$\Rightarrow \frac{\lambda+6}{\sqrt{\lambda^{2}+4 \lambda+44}} = 1$.
$\Rightarrow \lambda+6 = \sqrt{\lambda^{2}+4 \lambda+44}$.
Squaring both sides: $(\lambda+6)^{2} = \lambda^{2}+4 \lambda+44$.
$\Rightarrow \lambda^{2}+12 \lambda+36 = \lambda^{2}+4 \lambda+44$.
$\Rightarrow 8 \lambda = 8$.
$\Rightarrow \lambda = 1$.

Since $\vec{a}, \vec{b}$ and $\vec{c}$ are mutually perpendicular vectors,we have $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$.
It is given that $|\vec{a}| = |\vec{b}| = |\vec{c}|$.
Let the vector $\vec{a} + \vec{b} + \vec{c}$ be inclined to $\vec{a}, \vec{b}$ and $\vec{c}$ at angles $\theta_{1}, \theta_{2}$ and $\theta_{3}$ respectively.
Then,we have:
$\cos \theta_{1} = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} = \frac{\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{a} + \vec{c} \cdot \vec{a}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} = \frac{|\vec{a}|^2}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} = \frac{|\vec{a}|}{|\vec{a} + \vec{b} + \vec{c}|}$.
Similarly,$\cos \theta_{2} = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{b}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{b}|} = \frac{\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{b}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{b}|} = \frac{|\vec{b}|^2}{|\vec{a} + \vec{b} + \vec{c}| |\vec{b}|} = \frac{|\vec{b}|}{|\vec{a} + \vec{b} + \vec{c}|}$.
And $\cos \theta_{3} = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{c}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{c}|} = \frac{\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{c}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{c}|} = \frac{|\vec{c}|^2}{|\vec{a} + \vec{b} + \vec{c}| |\vec{c}|} = \frac{|\vec{c}|}{|\vec{a} + \vec{b} + \vec{c}|}$.
Since $|\vec{a}| = |\vec{b}| = |\vec{c}|$,we have $\cos \theta_{1} = \cos \theta_{2} = \cos \theta_{3}$,which implies $\theta_{1} = \theta_{2} = \theta_{3}$.
Hence,the vector $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}, \vec{b}$ and $\vec{c}$.

(A) Given the equation: $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}$
Using the distributive property of the dot product:
$\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = |\vec{a}|^{2} + |\vec{b}|^{2}$
Since $\vec{a} \cdot \vec{a} = |\vec{a}|^{2}$,$\vec{b} \cdot \vec{b} = |\vec{b}|^{2}$,and the dot product is commutative $(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a})$:
$|\vec{a}|^{2} + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^{2} = |\vec{a}|^{2} + |\vec{b}|^{2}$
Subtracting $|\vec{a}|^{2} + |\vec{b}|^{2}$ from both sides:
$2(\vec{a} \cdot \vec{b}) = 0$
$\vec{a} \cdot \vec{b} = 0$
Since $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0}$,the dot product of two non-zero vectors is zero if and only if they are perpendicular. Thus,$\vec{a} \perp \vec{b}$.

(C) Let $\theta$ be the angle between two vectors $\vec{a}$ and $\vec{b}$.
Assuming $\vec{a}$ and $\vec{b}$ are non-zero vectors,their magnitudes $|\vec{a}|$ and $|\vec{b}|$ are positive.
The dot product is defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$.
Given $\vec{a} \cdot \vec{b} \geq 0$,we have $|\vec{a}| |\vec{b}| \cos \theta \geq 0$.
Since $|\vec{a}| > 0$ and $|\vec{b}| > 0$,it follows that $\cos \theta \geq 0$.
For the angle $\theta$ between two vectors,we know $0 \leq \theta \leq \pi$.
Within this range,$\cos \theta \geq 0$ holds when $0 \leq \theta \leq \frac{\pi}{2}$.
Thus,$\vec{a} \cdot \vec{b} \geq 0$ when $0 \leq \theta \leq \frac{\pi}{2}$.
The correct answer is $C$.

(D) Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ be the angle between them.
Then,$|\vec{a}| = |\vec{b}| = 1$.
Now,$\vec{a}+\vec{b}$ is a unit vector if $|\vec{a}+\vec{b}| = 1$.
Squaring both sides,we get $|\vec{a}+\vec{b}|^2 = 1^2$.
$(\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = 1$.
$\vec{a} \cdot \vec{a} + 2(\vec{a} \cdot \vec{b}) + \vec{b} \cdot \vec{b} = 1$.
$|\vec{a}|^2 + 2|\vec{a}||\vec{b}| \cos \theta + |\vec{b}|^2 = 1$.
Since $|\vec{a}| = 1$ and $|\vec{b}| = 1$,we have $1^2 + 2(1)(1) \cos \theta + 1^2 = 1$.
$1 + 2 \cos \theta + 1 = 1$.
$2 \cos \theta = -1$.
$\cos \theta = -\frac{1}{2}$.
Since $\cos \theta = -\frac{1}{2}$,the angle $\theta = \frac{2\pi}{3}$.
Thus,the correct option is $D$.

(A) We know that for unit vectors $\hat{i}, \hat{j}, \hat{k}$ along the axes,the cross products are $\hat{j} \times \hat{k} = \hat{i}$,$\hat{i} \times \hat{k} = -\hat{j}$,and $\hat{i} \times \hat{j} = \hat{k}$.
Substituting these into the expression:
$\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})$
$= \hat{i} \cdot \hat{i} + \hat{j} \cdot (-\hat{j}) + \hat{k} \cdot \hat{k}$
$= 1 - \hat{j} \cdot \hat{j} + 1$
$= 1 - 1 + 1$
$= 1$
Thus,the correct answer is $A$.

(A) Let $\theta$ be the angle between two vectors $\vec{a}$ and $\vec{b}$.
Assuming $\vec{a}$ and $\vec{b}$ are non-zero vectors,we have $|\vec{a}| > 0$ and $|\vec{b}| > 0$.
The given condition is $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$.
Using the definitions of dot and cross products,we get $|\vec{a}||\vec{b}| |\cos \theta| = |\vec{a}||\vec{b}| |\sin \theta|$.
Since $|\vec{a}|$ and $|\vec{b}|$ are positive,we can divide both sides by $|\vec{a}||\vec{b}|$,yielding $|\cos \theta| = |\sin \theta|$.
This implies $|\tan \theta| = 1$.
For $\theta$ in the range $[0, \pi]$,$\tan \theta = 1$ gives $\theta = \frac{\pi}{4}$.
Thus,the condition holds when $\theta = \frac{\pi}{4}$.
The correct answer is $A$.

(D) Given vectors are $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$,and $\vec{c} = \hat{i} + 3\hat{j} - \hat{k}$.
First,calculate the vector $\vec{v} = \lambda\vec{b} + \vec{c}$:
$\vec{v} = \lambda(\hat{i} + \hat{j} - 2\hat{k}) + (\hat{i} + 3\hat{j} - \hat{k})$
$= (\lambda + 1)\hat{i} + (\lambda + 3)\hat{j} - (2\lambda + 1)\hat{k}$.
Since $\vec{a}$ is perpendicular to $\vec{v}$,their dot product must be zero,i.e.,$\vec{a} \cdot \vec{v} = 0$.
$(2\hat{i} - \hat{j} + \hat{k}) \cdot ((\lambda + 1)\hat{i} + (\lambda + 3)\hat{j} - (2\lambda + 1)\hat{k}) = 0$
$2(\lambda + 1) - 1(\lambda + 3) + 1(-2\lambda - 1) = 0$
$2\lambda + 2 - \lambda - 3 - 2\lambda - 1 = 0$
$-\lambda - 2 = 0$
$\lambda = -2$.

(N/A) Let $\widehat{OP}$ and $\widehat{OQ}$ be unit vectors making angles $A$ and $B$,respectively,with the positive direction of the $x$-axis. Then $\angle QOP = A-B$ (as shown in the figure).
We know that $\widehat{OP} = \cos A \hat{i} + \sin A \hat{j}$ and $\widehat{OQ} = \cos B \hat{i} + \sin B \hat{j}$.
By the definition of the dot product,$\widehat{OP} \cdot \widehat{OQ} = |\widehat{OP}| |\widehat{OQ}| \cos(A-B)$.
Since $\widehat{OP}$ and $\widehat{OQ}$ are unit vectors,$|\widehat{OP}| = 1$ and $|\widehat{OQ}| = 1$.
Therefore,$\widehat{OP} \cdot \widehat{OQ} = \cos(A-B) \quad \dots(1)$.
In terms of components,$\widehat{OP} \cdot \widehat{OQ} = (\cos A \hat{i} + \sin A \hat{j}) \cdot (\cos B \hat{i} + \sin B \hat{j})$.
Using the property $\hat{i} \cdot \hat{i} = 1, \hat{j} \cdot \hat{j} = 1, \hat{i} \cdot \hat{j} = 0$,we get:
$\widehat{OP} \cdot \widehat{OQ} = \cos A \cos B + \sin A \sin B \quad \dots(2)$.
From equations $(1)$ and $(2)$,we get:
$\cos(A-B) = \cos A \cos B + \sin A \sin B$.

(A) Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = 3\hat{i} + 4\hat{j} - \hat{k}$.
The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
First,calculate the dot product: $\vec{a} \cdot \vec{b} = (2)(3) + (-1)(4) + (1)(-1) = 6 - 4 - 1 = 1$.
Next,calculate the magnitudes: $|\vec{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $|\vec{b}| = \sqrt{3^2 + 4^2 + (-1)^2} = \sqrt{9 + 16 + 1} = \sqrt{26}$.
Substitute these values into the formula: $\cos \theta = \frac{1}{\sqrt{6} \cdot \sqrt{26}} = \frac{1}{\sqrt{156}} = \frac{1}{\sqrt{4 \times 39}} = \frac{1}{2\sqrt{39}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{1}{2\sqrt{39}}\right)$.

(A) Given vectors are $\vec{a}=3 \hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}-2 \hat{j}+4 \hat{k}$.
First,calculate the dot product $\vec{a} \cdot \vec{b} = (3)(2) + (1)(-2) + (2)(4) = 6 - 2 + 8 = 12$.
Next,calculate the magnitudes $|\vec{a}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9+1+4} = \sqrt{14}$ and $|\vec{b}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4+4+16} = \sqrt{24} = 2\sqrt{6}$.
Using the formula $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$,we get $\cos \theta = \frac{12}{\sqrt{14} \cdot 2\sqrt{6}} = \frac{6}{\sqrt{84}} = \frac{6}{2\sqrt{21}} = \frac{3}{\sqrt{21}}$.
Now,use the identity $\sin \theta = \sqrt{1 - \cos^2 \theta}$.
$\sin \theta = \sqrt{1 - (\frac{3}{\sqrt{21}})^2} = \sqrt{1 - \frac{9}{21}} = \sqrt{\frac{12}{21}} = \sqrt{\frac{4}{7}} = \frac{2}{\sqrt{7}}$.

(D) Given position vectors are $\overrightarrow{OA} = \hat{i}-\hat{j}+\hat{k}$,$\overrightarrow{OB} = 2\hat{i}-\hat{j}+3\hat{k}$,$\overrightarrow{OC} = 2\hat{i}-3\hat{k}$,and $\overrightarrow{OD} = 3\hat{i}-2\hat{j}+\hat{k}$.
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2-1)\hat{i} + (-1 - (-1))\hat{j} + (3-1)\hat{k} = \hat{i} + 0\hat{j} + 2\hat{k}$.
$\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} = (3-2)\hat{i} + (-2-0)\hat{j} + (1 - (-3))\hat{k} = \hat{i} - 2\hat{j} + 4\hat{k}$.
The projection of $\overrightarrow{AB}$ on $\overrightarrow{CD}$ is given by $\frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{CD}|}$.
$\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(1) + (0)(-2) + (2)(4) = 1 + 0 + 8 = 9$.
$|\overrightarrow{CD}| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$.
Therefore,the projection is $\frac{9}{\sqrt{21}} = \frac{9\sqrt{21}}{21} = \frac{3\sqrt{21}}{7}$ units.

(A) Given $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{r} \times \overrightarrow{b}$,we have $\overrightarrow{r} \times (\overrightarrow{a} - \overrightarrow{b}) = 0$.
This implies $\overrightarrow{r}$ is parallel to $(\overrightarrow{a} - \overrightarrow{b})$.
Calculate $\overrightarrow{a} - \overrightarrow{b} = (2-7)\hat{i} + (-3-1)\hat{j} + (4 - (-6))\hat{k} = -5\hat{i} - 4\hat{j} + 10\hat{k}$.
So,$\overrightarrow{r} = \lambda(-5\hat{i} - 4\hat{j} + 10\hat{k})$.
Given $\overrightarrow{r} \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$,substitute $\overrightarrow{r}$:
$\lambda(-5\hat{i} - 4\hat{j} + 10\hat{k}) \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$.
$\lambda(-5 - 8 + 10) = -3 \Rightarrow -3\lambda = -3 \Rightarrow \lambda = 1$.
Thus,$\overrightarrow{r} = -5\hat{i} - 4\hat{j} + 10\hat{k}$.
Finally,calculate $\overrightarrow{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k}) = (-5)(2) + (-4)(-3) + (10)(1) = -10 + 12 + 10 = 12$.

(N/A) Let $ABCD$ be a parallelogram such that $\overrightarrow{AB} = \vec{p}$ and $\overrightarrow{AD} = \vec{q}$. Then $\overrightarrow{BC} = \vec{q}$.
By the triangle law of vector addition,we have:
$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \vec{p} + \vec{q} = \vec{a} \quad \dots(i)$
$\overrightarrow{BD} = \overrightarrow{AD} - \overrightarrow{AB} = \vec{q} - \vec{p} = \vec{b} \quad \dots(ii)$
Adding $(i)$ and $(ii)$,we get $\vec{a} + \vec{b} = 2\vec{q} \Rightarrow \vec{q} = \frac{1}{2}(\vec{a} + \vec{b})$.
Subtracting $(ii)$ from $(i)$,we get $\vec{a} - \vec{b} = 2\vec{p} \Rightarrow \vec{p} = \frac{1}{2}(\vec{a} - \vec{b})$.
The area of the parallelogram is given by $|\vec{p} \times \vec{q}|$.
$|\vec{p} \times \vec{q}| = |\frac{1}{2}(\vec{a} - \vec{b}) \times \frac{1}{2}(\vec{a} + \vec{b})| = \frac{1}{4} |\vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}|$.
Since $\vec{a} \times \vec{a} = 0$ and $\vec{b} \times \vec{b} = 0$,and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$,we have:
$|\vec{p} \times \vec{q}| = \frac{1}{4} |\vec{a} \times \vec{b} + \vec{a} \times \vec{b}| = \frac{1}{4} |2(\vec{a} \times \vec{b})| = \frac{1}{2} |\vec{a} \times \vec{b}|$.
Now,for diagonals $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} - \hat{k}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix} = \hat{i}(1 - 3) - \hat{j}(-2 - 1) + \hat{k}(6 + 1) = -2\hat{i} + 3\hat{j} + 7\hat{k}$.
Area $= \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{1}{2} \sqrt{(-2)^2 + 3^2 + 7^2} = \frac{1}{2} \sqrt{4 + 9 + 49} = \frac{\sqrt{62}}{2} \text{ sq units}$.

(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
Expanding the given equation $|\vec{a}-\vec{b}|^{2}+|\vec{a}-\vec{c}|^{2}=8$:
$(|\vec{a}|^{2} + |\vec{b}|^{2} - 2\vec{a}\cdot\vec{b}) + (|\vec{a}|^{2} + |\vec{c}|^{2} - 2\vec{a}\cdot\vec{c}) = 8$
Since $|\vec{a}|^{2} = |\vec{b}|^{2} = |\vec{c}|^{2} = 1$,we have:
$(1 + 1 - 2\vec{a}\cdot\vec{b}) + (1 + 1 - 2\vec{a}\cdot\vec{c}) = 8$
$4 - 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c}) = 8$
$-2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c}) = 4$
$\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = -2$
Now,evaluate $|\vec{a}+2\vec{b}|^{2}+|\vec{a}+2\vec{c}|^{2}$:
$= (|\vec{a}|^{2} + 4|\vec{b}|^{2} + 4\vec{a}\cdot\vec{b}) + (|\vec{a}|^{2} + 4|\vec{c}|^{2} + 4\vec{a}\cdot\vec{c})$
$= (1 + 4 + 4\vec{a}\cdot\vec{b}) + (1 + 4 + 4\vec{a}\cdot\vec{c})$
$= 10 + 4(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c})$
Substitute the value $\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = -2$:
$= 10 + 4(-2) = 10 - 8 = 2$.

(C) Using the section formula,the position vector of point $P$ is given by:
$\overrightarrow{OP} = \frac{\lambda(2\hat{i}+\hat{j}+3\hat{k}) + 1(\hat{i}+\hat{j}+\hat{k})}{\lambda+1} = \frac{2\lambda+1}{\lambda+1}\hat{i} + \frac{\lambda+1}{\lambda+1}\hat{j} + \frac{3\lambda+1}{\lambda+1}\hat{k}$
Now,calculate $\overrightarrow{OB} \cdot \overrightarrow{OP}$:
$\overrightarrow{OB} \cdot \overrightarrow{OP} = (2\hat{i}+\hat{j}+3\hat{k}) \cdot \left( \frac{2\lambda+1}{\lambda+1}\hat{i} + \hat{j} + \frac{3\lambda+1}{\lambda+1}\hat{k} \right) = \frac{4\lambda+2 + \lambda+1 + 9\lambda+3}{\lambda+1} = \frac{14\lambda+6}{\lambda+1}$
Next,calculate $\overrightarrow{OA} \times \overrightarrow{OP}$:
$\overrightarrow{OA} \times \overrightarrow{OP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ \frac{2\lambda+1}{\lambda+1} & 1 & \frac{3\lambda+1}{\lambda+1} \end{vmatrix} = \left( \frac{3\lambda+1}{\lambda+1} - 1 \right)\hat{i} - \left( \frac{3\lambda+1}{\lambda+1} - \frac{2\lambda+1}{\lambda+1} \right)\hat{j} + \left( 1 - \frac{2\lambda+1}{\lambda+1} \right)\hat{k}$
$= \frac{2\lambda}{\lambda+1}\hat{i} - \frac{\lambda}{\lambda+1}\hat{j} - \frac{\lambda}{\lambda+1}\hat{k}$
$|\overrightarrow{OA} \times \overrightarrow{OP}|^{2} = \frac{4\lambda^{2} + \lambda^{2} + \lambda^{2}}{(\lambda+1)^{2}} = \frac{6\lambda^{2}}{(\lambda+1)^{2}}$
Substituting these into the given equation:
$\frac{14\lambda+6}{\lambda+1} - 3 \left( \frac{6\lambda^{2}}{(\lambda+1)^{2}} \right) = 6$
Multiply by $(\lambda+1)^{2}$:
$(14\lambda+6)(\lambda+1) - 18\lambda^{2} = 6(\lambda+1)^{2}$
$14\lambda^{2} + 20\lambda + 6 - 18\lambda^{2} = 6(\lambda^{2} + 2\lambda + 1)$
$-4\lambda^{2} + 20\lambda + 6 = 6\lambda^{2} + 12\lambda + 6$
$10\lambda^{2} - 8\lambda = 0$
Since $\lambda > 0$,we have $10\lambda = 8 \Rightarrow \lambda = 0.8$.

(D) Let $\vec{p} = a \hat{i} + b \hat{j} + c \hat{k}$ and $\vec{q} = b \hat{i} + c \hat{j} + a \hat{k}$.
Given $a^{2} + b^{2} + c^{2} = 1$,we have $|\vec{p}| = \sqrt{a^{2} + b^{2} + c^{2}} = 1$ and $|\vec{q}| = \sqrt{b^{2} + c^{2} + a^{2}} = 1$.
The dot product is $\vec{p} \cdot \vec{q} = ab + bc + ca$.
Let $a \cos \theta = b \cos \left(\theta + \frac{2\pi}{3}\right) = c \cos \left(\theta + \frac{4\pi}{3}\right) = k$.
Then $a = \frac{k}{\cos \theta}$,$b = \frac{k}{\cos(\theta + 2\pi/3)}$,$c = \frac{k}{\cos(\theta + 4\pi/3)}$.
Since $a+b+c = k(\sec \theta + \sec(\theta + 2\pi/3) + \sec(\theta + 4\pi/3)) = 0$ (using the identity for sum of secants),we have $(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) = 0$.
Since $a^{2} + b^{2} + c^{2} = 1$,we get $1 + 2(ab + bc + ca) = 0$,so $ab + bc + ca = -1/2$.
Thus,$\cos \phi = \frac{\vec{p} \cdot \vec{q}}{|\vec{p}||\vec{q}|} = \frac{-1/2}{1 \times 1} = -1/2$.
Therefore,$\phi = \frac{2\pi}{3}$.

(A) Given that the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is equal to the projection of $\overrightarrow{c}$ on $\overrightarrow{a}$.
$\Rightarrow \frac{\overrightarrow{b} \cdot \overrightarrow{a}}{|\overrightarrow{a}|} = \frac{\overrightarrow{c} \cdot \overrightarrow{a}}{|\overrightarrow{a}|} \Rightarrow \overrightarrow{b} \cdot \overrightarrow{a} = \overrightarrow{c} \cdot \overrightarrow{a}$.
Also,$\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$,so $\overrightarrow{b} \cdot \overrightarrow{c} = 0$.
Let $k = |\overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c}|$.
Squaring both sides,we get:
$k^2 = |\overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b}) - 2(\overrightarrow{a} \cdot \overrightarrow{c}) - 2(\overrightarrow{b} \cdot \overrightarrow{c})$.
Since $\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{a} \cdot \overrightarrow{c}$ and $\overrightarrow{b} \cdot \overrightarrow{c} = 0$,the terms cancel out:
$k^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b} - \overrightarrow{a} \cdot \overrightarrow{c}) - 2(0) = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2$.
Substituting the given magnitudes $|\overrightarrow{a}|=2, |\overrightarrow{b}|=4, |\overrightarrow{c}|=4$:
$k^2 = 2^2 + 4^2 + 4^2 = 4 + 16 + 16 = 36$.
Therefore,$k = \sqrt{36} = 6$.

(A) Given $|\overrightarrow{x}+\overrightarrow{y}|=|\overrightarrow{x}|$. Squaring both sides,we get:
$|\overrightarrow{x}|^2+|\overrightarrow{y}|^2+2\overrightarrow{x}\cdot\overrightarrow{y}=|\overrightarrow{x}|^2$
$|\overrightarrow{y}|^2+2\overrightarrow{x}\cdot\overrightarrow{y}=0$ --- $(1)$
Given that $(2\overrightarrow{x}+\lambda\overrightarrow{y})$ is perpendicular to $\overrightarrow{y}$,their dot product is zero:
$(2\overrightarrow{x}+\lambda\overrightarrow{y})\cdot\overrightarrow{y}=0$
$2\overrightarrow{x}\cdot\overrightarrow{y}+\lambda|\overrightarrow{y}|^2=0$ --- $(2)$
From equation $(1)$,we have $2\overrightarrow{x}\cdot\overrightarrow{y}=-|\overrightarrow{y}|^2$. Substituting this into equation $(2)$:
$-|\overrightarrow{y}|^2+\lambda|\overrightarrow{y}|^2=0$
$(\lambda-1)|\overrightarrow{y}|^2=0$
Since $\overrightarrow{y}$ is a non-zero vector,$|\overrightarrow{y}|^2 \neq 0$. Therefore,$\lambda-1=0$,which gives $\lambda=1$.

(D) Let $\theta$ be the angle between the unit vectors $\overrightarrow{a}$ and $\overrightarrow{b}$.
We know that $|\overrightarrow{a}+\overrightarrow{b}| = \sqrt{|\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + 2\overrightarrow{a} \cdot \overrightarrow{b}} = \sqrt{1+1+2\cos \theta} = \sqrt{2+2\cos \theta} = 2|\cos(\theta/2)|$.
Similarly,$|\overrightarrow{a}-\overrightarrow{b}| = \sqrt{|\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 - 2\overrightarrow{a} \cdot \overrightarrow{b}} = \sqrt{2-2\cos \theta} = 2|\sin(\theta/2)|$.
The expression becomes $f(\theta) = \sqrt{3}(2|\cos(\theta/2)|) + 2|\sin(\theta/2)|$.
Using the Cauchy-Schwarz inequality or the property $a\cos x + b\sin x \leq \sqrt{a^2+b^2}$,the maximum value is $\sqrt{(\sqrt{3} \times 2)^2 + 2^2} = \sqrt{12+4} = \sqrt{16} = 4$.

(B) Given $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{b} \times \overrightarrow{r}$,we can write $\overrightarrow{r} \times \overrightarrow{a} + \overrightarrow{r} \times \overrightarrow{b} = 0$,which implies $\overrightarrow{r} \times (\overrightarrow{a} + \overrightarrow{b}) = 0$.
This means $\overrightarrow{r}$ is parallel to $(\overrightarrow{a} + \overrightarrow{b})$.
$\overrightarrow{a} + \overrightarrow{b} = (\hat{i} + 2\hat{j} - 3\hat{k}) + (2\hat{i} - 3\hat{j} + 5\hat{k}) = 3\hat{i} - \hat{j} + 2\hat{k}$.
So,$\overrightarrow{r} = \lambda(3\hat{i} - \hat{j} + 2\hat{k})$ for some scalar $\lambda$.
Substituting $\overrightarrow{r}$ into $\overrightarrow{r} \cdot (\alpha\hat{i} + 2\hat{j} + \hat{k}) = 3$:
$\lambda(3\alpha - 2 + 2) = 3 \Rightarrow 3\lambda\alpha = 3 \Rightarrow \lambda\alpha = 1$.
Substituting $\overrightarrow{r}$ into $\overrightarrow{r} \cdot (2\hat{i} + 5\hat{j} - \alpha\hat{k}) = -1$:
$\lambda(6 - 5 - 2\alpha) = -1 \Rightarrow \lambda(1 - 2\alpha) = -1 \Rightarrow \lambda - 2\lambda\alpha = -1$.
Since $\lambda\alpha = 1$,we have $\lambda - 2(1) = -1 \Rightarrow \lambda = 1$.
Then $\alpha = 1/\lambda = 1$.
Thus,$\overrightarrow{r} = 3\hat{i} - \hat{j} + 2\hat{k}$.
$|\overrightarrow{r}|^{2} = 3^{2} + (-1)^{2} + 2^{2} = 9 + 1 + 4 = 14$.
Finally,$\alpha + |\overrightarrow{r}|^{2} = 1 + 14 = 15$.

(C) Since $\overrightarrow{x}$ lies in the plane of $\overrightarrow{a}$ and $\overrightarrow{b}$,we can write $\overrightarrow{x} = \lambda\overrightarrow{a} + \mu\overrightarrow{b} = \lambda(2\hat{i} - \hat{j} + \hat{k}) + \mu(\hat{i} + 2\hat{j} - \hat{k}) = (2\lambda + \mu)\hat{i} + (2\mu - \lambda)\hat{j} + (\lambda - \mu)\hat{k}$.
Given $\overrightarrow{x} \perp (3\hat{i} + 2\hat{j} - \hat{k})$,so $\overrightarrow{x} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 0$.
$3(2\lambda + \mu) + 2(2\mu - \lambda) - 1(\lambda - \mu) = 0 \implies 6\lambda + 3\mu + 4\mu - 2\lambda - \lambda + \mu = 0 \implies 3\lambda + 8\mu = 0 \implies \lambda = -\frac{8}{3}\mu$.
Projection of $\overrightarrow{x}$ on $\overrightarrow{a}$ is $\frac{\overrightarrow{x} \cdot \overrightarrow{a}}{|\overrightarrow{a}|} = \frac{17\sqrt{6}}{2}$.
$|\overrightarrow{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$.
$\overrightarrow{x} \cdot \overrightarrow{a} = (2\lambda + \mu)(2) + (2\mu - \lambda)(-1) + (\lambda - \mu)(1) = 4\lambda + 2\mu - 2\mu + \lambda + \lambda - \mu = 6\lambda - \mu$.
So,$\frac{6\lambda - \mu}{\sqrt{6}} = \frac{17\sqrt{6}}{2} \implies 6\lambda - \mu = 51$.
Substituting $\lambda = -\frac{8}{3}\mu$ into $6\lambda - \mu = 51$: $6(-\frac{8}{3}\mu) - \mu = 51 \implies -16\mu - \mu = 51 \implies -17\mu = 51 \implies \mu = -3$.
Then $\lambda = -\frac{8}{3}(-3) = 8$.
Thus,$\overrightarrow{x} = 8(2\hat{i} - \hat{j} + \hat{k}) - 3(\hat{i} + 2\hat{j} - \hat{k}) = (16-3)\hat{i} + (-8-6)\hat{j} + (8+3)\hat{k} = 13\hat{i} - 14\hat{j} + 11\hat{k}$.
$|\overrightarrow{x}|^{2} = 13^2 + (-14)^2 + 11^2 = 169 + 196 + 121 = 486$.

(B) Let $\vec{c} = \overline{AB},$ $\vec{b} = \overline{AC},$ and $\vec{a} = \overline{BC}.$ Given $|\vec{a}|=8, |\vec{b}|=7, |\vec{c}|=10.$
Using the Law of Cosines in $\triangle ABC$ at vertex $A$ where $\theta$ is the angle between $\overline{AB}$ and $\overline{AC}$:
$|\vec{a}|^2 = |\vec{b}|^2 + |\vec{c}|^2 - 2|\vec{b}||\vec{c}| \cos \theta$
$8^2 = 7^2 + 10^2 - 2(7)(10) \cos \theta$
$64 = 49 + 100 - 140 \cos \theta$
$140 \cos \theta = 149 - 64 = 85$
$\cos \theta = \frac{85}{140} = \frac{17}{28}.$
The projection of vector $\overline{AB}$ (which is $\vec{c}$) on $\overline{AC}$ (which is $\vec{b}$) is given by $|\vec{c}| \cos \theta.$
Projection $= 10 \times \frac{17}{28} = \frac{170}{28} = \frac{85}{14}.$

(A) Since $\overrightarrow{c}$ is coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$,we can write $\overrightarrow{c} = x\overrightarrow{a} + y\overrightarrow{b}$.
Given $\overrightarrow{b} \cdot \overrightarrow{c} = 0$,we have $\overrightarrow{b} \cdot (x\overrightarrow{a} + y\overrightarrow{b}) = 0$,which implies $x(\overrightarrow{a} \cdot \overrightarrow{b}) + y|\overrightarrow{b}|^{2} = 0$.
Calculating $\overrightarrow{a} \cdot \overrightarrow{b} = (-1)(2) + (1)(0) + (1)(1) = -1$ and $|\overrightarrow{b}|^{2} = 2^{2} + 0^{2} + 1^{2} = 5$.
So,$-x + 5y = 0 \Rightarrow x = 5y$.
Thus,$\overrightarrow{c} = y(5\overrightarrow{a} + \overrightarrow{b}) = y(5(-\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + \hat{k})) = y(-3\hat{i} + 5\hat{j} + 6\hat{k})$.
Given $\overrightarrow{a} \cdot \overrightarrow{c} = 7$,we have $y((-1)(-3) + (1)(5) + (1)(6)) = 7 \Rightarrow y(3 + 5 + 6) = 7 \Rightarrow 14y = 7 \Rightarrow y = \frac{1}{2}$.
So,$\overrightarrow{c} = -\frac{3}{2}\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}$.
Then $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = (-\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + \hat{k}) + (-\frac{3}{2}\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}) = (2 - 1 - \frac{3}{2})\hat{i} + (1 + \frac{5}{2})\hat{j} + (1 + 1 + 3)\hat{k} = -\frac{1}{2}\hat{i} + \frac{7}{2}\hat{j} + 5\hat{k}$.
Finally,$2|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}|^{2} = 2((\frac{-1}{2})^{2} + (\frac{7}{2})^{2} + 5^{2}) = 2(\frac{1}{4} + \frac{49}{4} + 25) = 2(\frac{50}{4} + 25) = 2(12.5 + 25) = 2(37.5) = 75$.

(C) Given $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{c} \times \overrightarrow{a}$,we can write $(\overrightarrow{r} - \overrightarrow{c}) \times \overrightarrow{a} = 0$.
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 $\overrightarrow{r}$:
$(\overrightarrow{c} + \lambda \overrightarrow{a}) \cdot \overrightarrow{b} = 0 \Rightarrow \overrightarrow{c} \cdot \overrightarrow{b} + \lambda (\overrightarrow{a} \cdot \overrightarrow{b}) = 0$.
Calculating the dot products:
$\overrightarrow{c} \cdot \overrightarrow{b} = (1)(1) + (-1)(-1) + (-1)(0) = 1 + 1 = 2$.
$\overrightarrow{a} \cdot \overrightarrow{b} = (1)(1) + (2)(-1) + (-1)(0) = 1 - 2 = -1$.
Thus,$2 + \lambda(-1) = 0 \Rightarrow \lambda = 2$.
Now,$\overrightarrow{r} \cdot \overrightarrow{a} = (\overrightarrow{c} + 2\overrightarrow{a}) \cdot \overrightarrow{a} = \overrightarrow{c} \cdot \overrightarrow{a} + 2|\overrightarrow{a}|^2$.
$\overrightarrow{c} \cdot \overrightarrow{a} = (1)(1) + (-1)(2) + (-1)(-1) = 1 - 2 + 1 = 0$.
$|\overrightarrow{a}|^2 = 1^2 + 2^2 + (-1)^2 = 1 + 4 + 1 = 6$.
Therefore,$\overrightarrow{r} \cdot \overrightarrow{a} = 0 + 2(6) = 12$.

(D) Let the vertices of the floor be $A(0,0,0)$,$B(10,0,0)$,$C(10,10,0)$,and $D(0,10,0)$. Let the height of the hall be $h$. Then the vertices of the ceiling are $E(0,0,h)$,$F(10,0,h)$,$G(10,10,h)$,and $H(0,10,h)$.
The vector $\overrightarrow{AG} = (10-0)\hat{i} + (10-0)\hat{j} + (h-0)\hat{k} = 10\hat{i} + 10\hat{j} + h\hat{k}$.
The vector $\overrightarrow{BH} = (0-10)\hat{i} + (10-0)\hat{j} + (h-0)\hat{k} = -10\hat{i} + 10\hat{j} + h\hat{k}$.
The angle $\theta$ between $\overrightarrow{AG}$ and $\overrightarrow{BH}$ is given by $\cos \theta = \frac{\overrightarrow{AG} \cdot \overrightarrow{BH}}{|\overrightarrow{AG}| |\overrightarrow{BH}|}$.
Given $\cos \theta = \frac{1}{5}$,we have:
$\frac{1}{5} = \frac{(10\hat{i} + 10\hat{j} + h\hat{k}) \cdot (-10\hat{i} + 10\hat{j} + h\hat{k})}{\sqrt{10^2 + 10^2 + h^2} \sqrt{(-10)^2 + 10^2 + h^2}}$
$\frac{1}{5} = \frac{-100 + 100 + h^2}{200 + h^2} = \frac{h^2}{200 + h^2}$.
$200 + h^2 = 5h^2 \Rightarrow 4h^2 = 200 \Rightarrow h^2 = 50 \Rightarrow h = \sqrt{50} = 5\sqrt{2}$ meters.

(D) Let $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$.
Let $\vec{v_1} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{v_2} = -\lambda\hat{i} + 2\hat{j} + 3\hat{k}$.
The sum of the vectors is $\vec{b} = \vec{v_1} + \vec{v_2} = (2 - \lambda)\hat{i} + 6\hat{j} - 2\hat{k}$.
The projection of $\vec{a}$ on $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = 1$.
First,calculate the dot product: $\vec{a} \cdot \vec{b} = (1)(2 - \lambda) + (2)(6) + (1)(-2) = 2 - \lambda + 12 - 2 = 12 - \lambda$.
Next,calculate the magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{(2 - \lambda)^2 + 6^2 + (-2)^2} = \sqrt{4 - 4\lambda + \lambda^2 + 36 + 4} = \sqrt{\lambda^2 - 4\lambda + 44}$.
Setting the projection equal to $1$: $\frac{12 - \lambda}{\sqrt{\lambda^2 - 4\lambda + 44}} = 1$.
Squaring both sides: $(12 - \lambda)^2 = \lambda^2 - 4\lambda + 44$.
$144 - 24\lambda + \lambda^2 = \lambda^2 - 4\lambda + 44$.
$144 - 44 = 24\lambda - 4\lambda$.
$100 = 20\lambda$.
$\lambda = 5$.

(C) Given $|2 \vec{a}+3 \vec{b}|=|3 \vec{a}+\vec{b}|$.
Squaring both sides,we get $|2 \vec{a}+3 \vec{b}|^{2}=|3 \vec{a}+\vec{b}|^{2}$.
$(2 \vec{a}+3 \vec{b}) \cdot (2 \vec{a}+3 \vec{b}) = (3 \vec{a}+\vec{b}) \cdot (3 \vec{a}+\vec{b})$.
$4|\vec{a}|^{2} + 12(\vec{a} \cdot \vec{b}) + 9|\vec{b}|^{2} = 9|\vec{a}|^{2} + 6(\vec{a} \cdot \vec{b}) + |\vec{b}|^{2}$.
Rearranging the terms,we get $8|\vec{b}|^{2} - 5|\vec{a}|^{2} + 6(\vec{a} \cdot \vec{b}) = 0$.
Since $\frac{1}{8} \vec{a}$ is a unit vector,$|\frac{1}{8} \vec{a}| = 1 \Rightarrow |\vec{a}| = 8$.
Also,$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 60^{\circ} = 8 \cdot |\vec{b}| \cdot \frac{1}{2} = 4|\vec{b}|$.
Substituting these values into the equation: $8|\vec{b}|^{2} + 6(4|\vec{b}|) - 5(8)^{2} = 0$.
$8|\vec{b}|^{2} + 24|\vec{b}| - 320 = 0$.
Dividing by $8$,we get $|\vec{b}|^{2} + 3|\vec{b}| - 40 = 0$.
$(|\vec{b}| + 8)(|\vec{b}| - 5) = 0$.
Since the magnitude $|\vec{b}|$ must be positive,$|\vec{b}| = 5$.

(D) Let the magnitude of the vectors be $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$. Since they are mutually perpendicular,$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$.
Let $\vec{v} = \vec{a} + \vec{b} + \vec{c}$. Then $|\vec{v}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 3k^2$.
Thus,$|\vec{v}| = \sqrt{3}k$.
The angle $\theta$ between $\vec{a}$ and $\vec{v}$ is given by $\cos \theta = \frac{\vec{a} \cdot \vec{v}}{|\vec{a}| |\vec{v}|} = \frac{\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c})}{k \cdot \sqrt{3}k} = \frac{|\vec{a}|^2 + 0 + 0}{\sqrt{3}k^2} = \frac{k^2}{\sqrt{3}k^2} = \frac{1}{\sqrt{3}}$.
We need to find $36 \cos^2 2\theta$. Using the identity $\cos 2\theta = 2 \cos^2 \theta - 1$,we get $\cos 2\theta = 2(\frac{1}{\sqrt{3}})^2 - 1 = 2(\frac{1}{3}) - 1 = \frac{2}{3} - 1 = -\frac{1}{3}$.
Therefore,$36 \cos^2 2\theta = 36(-\frac{1}{3})^2 = 36(\frac{1}{9}) = 4$.

(A) Let $\vec{a} = \overrightarrow{BC}$,$\vec{b} = \overrightarrow{AC}$,and $\vec{c} = \overrightarrow{BA}$.
Given $|\vec{a}| = 3$,$|\vec{b}| = 5$,and $|\vec{c}| = 7$.
In $\triangle ABC$,by the Law of Cosines at vertex $B$:
$|\overrightarrow{AC}|^2 = |\overrightarrow{BA}|^2 + |\overrightarrow{BC}|^2 - 2 |\overrightarrow{BA}| |\overrightarrow{BC}| \cos(\angle ABC)$
$5^2 = 7^2 + 3^2 - 2(7)(3) \cos(\angle ABC)$
$25 = 49 + 9 - 42 \cos(\angle ABC)$
$25 = 58 - 42 \cos(\angle ABC)$
$42 \cos(\angle ABC) = 58 - 25 = 33$
$\cos(\angle ABC) = \frac{33}{42} = \frac{11}{14}$.
The projection of vector $\overrightarrow{BA}$ on $\overrightarrow{BC}$ is given by $|\overrightarrow{BA}| \cos(\angle ABC)$.
Projection $= 7 \times \frac{11}{14} = \frac{11}{2}$.

(A) Given $\vec{v}_{1} = \sqrt{3} p \hat{i} + \hat{j}$ and $\vec{v}_{2} = 2 \hat{i} + (p + 1) \hat{j}$.
Since rotation preserves the magnitude of the vector,$|\vec{v}_{1}| = |\vec{v}_{2}|$.
$(\sqrt{3}p)^2 + 1^2 = 2^2 + (p + 1)^2$
$3p^2 + 1 = 4 + p^2 + 2p + 1$
$2p^2 - 2p - 4 = 0 \Rightarrow p^2 - p - 2 = 0$.
Solving for $p$,$(p - 2)(p + 1) = 0$. Since $p > 0$,we have $p = 2$.
Thus,$\vec{v}_{1} = 2\sqrt{3} \hat{i} + \hat{j}$ and $\vec{v}_{2} = 2 \hat{i} + 3 \hat{j}$.
$|\vec{v}_{1}| = \sqrt{(2\sqrt{3})^2 + 1^2} = \sqrt{12 + 1} = \sqrt{13}$.
$|\vec{v}_{2}| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$.
Using $\vec{v}_{1} \cdot \vec{v}_{2} = |\vec{v}_{1}| |\vec{v}_{2}| \cos \theta$,we get $(2\sqrt{3})(2) + (1)(3) = \sqrt{13} \cdot \sqrt{13} \cos \theta$.
$4\sqrt{3} + 3 = 13 \cos \theta \Rightarrow \cos \theta = \frac{4\sqrt{3} + 3}{13}$.
Since $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{(4\sqrt{3} + 3)^2}{169} = \frac{169 - (48 + 9 + 24\sqrt{3})}{169} = \frac{112 - 24\sqrt{3}}{169}$.
$\sin \theta = \frac{\sqrt{112 - 24\sqrt{3}}}{13} = \frac{\sqrt{4(28 - 6\sqrt{3})}}{13} = \frac{2\sqrt{(3\sqrt{3} - 1)^2}}{13} = \frac{2(3\sqrt{3} - 1)}{13} = \frac{6\sqrt{3} - 2}{13}$.
$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{6\sqrt{3} - 2}{4\sqrt{3} + 3}$.
Comparing with $\frac{\alpha \sqrt{3} - 2}{4\sqrt{3} + 3}$,we get $\alpha = 6$.

(C) Given: $|\vec{a}|=2$ and $|\vec{b}|=5$.
We know that $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta = 8$.
Substituting the values: $2 \times 5 \times \sin \theta = 8 \implies 10 \sin \theta = 8 \implies \sin \theta = \frac{8}{10} = \frac{4}{5}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\cos^2 \theta = 1 - \sin^2 \theta = 1 - (\frac{4}{5})^2 = 1 - \frac{16}{25} = \frac{9}{25}$.
Thus,$|\cos \theta| = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
Now,$|\vec{a} \cdot \vec{b}| = |\vec{a}||\vec{b}| |\cos \theta|$.
$|\vec{a} \cdot \vec{b}| = 2 \times 5 \times \frac{3}{5} = 6$.