MHT CET 2022 Mathematics Question Paper with Answer and Solution

(D) For the line to lie on the plane,every point on the line must satisfy the equation of the plane.
Since the line passes through the point $(4, 2, k)$,this point must satisfy the plane equation $2x - 4y + z = 7$.
Substituting the coordinates of the point into the plane equation:
$2(4) - 4(2) + k = 7$
$8 - 8 + k = 7$
$k = 7$
Thus,the value of $k$ is $7$.

(A) Given point $P = (3, 2, 6)$.
Point $Q$ on the line is $(1 - 3\mu, -1 + \mu, 2 + 5\mu)$.
The vector $\vec{PQ} = (1 - 3\mu - 3)\hat{i} + (-1 + \mu - 2)\hat{j} + (2 + 5\mu - 6)\hat{k} = (-2 - 3\mu)\hat{i} + (-3 + \mu)\hat{j} + (-4 + 5\mu)\hat{k}$.
The normal vector to the plane $x - 4y + 3z = 1$ is $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$.
Since $\vec{PQ}$ is parallel to the plane,$\vec{PQ} \cdot \vec{n} = 0$.
$(-2 - 3\mu)(1) + (-3 + \mu)(-4) + (-4 + 5\mu)(3) = 0$.
$-2 - 3\mu + 12 - 4\mu - 12 + 15\mu = 0$.
$8\mu - 2 = 0$.
$8\mu = 2$.
$\mu = \frac{2}{8} = \frac{1}{4}$.

(C) For the four points to be coplanar,the determinant of the vectors formed by them must be zero. Let the points be $A(-\lambda^2, 1, 1)$,$B(1, -\lambda^2, 1)$,$C(1, 1, -\lambda^2)$,and $D(-1, -1, 1)$.
The vectors $\vec{AB} = (1+\lambda^2, -\lambda^2-1, 0)$,$\vec{AC} = (1+\lambda^2, 0, -\lambda^2-1)$,and $\vec{AD} = (-1+\lambda^2, -2, 0)$ must be coplanar.
Alternatively,using the condition that the four points $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3), (x_4, y_4, z_4)$ are coplanar:
$\begin{vmatrix} x_1-x_4 & y_1-y_4 & z_1-z_4 \\ x_2-x_4 & y_2-y_4 & z_2-z_4 \\ x_3-x_4 & y_3-y_4 & z_3-z_4 \end{vmatrix} = 0$
Substituting the points:
$\begin{vmatrix} -\lambda^2+1 & 2 & 0 \\ 2 & -\lambda^2+1 & 0 \\ 2 & 2 & -\lambda^2-1 \end{vmatrix} = 0$
Expanding along the third column:
$(-\lambda^2-1) [(-\lambda^2+1)^2 - 4] = 0$
$(-\lambda^2-1) [(\lambda^4 - 2\lambda^2 + 1) - 4] = 0$
$(-\lambda^2-1) (\lambda^4 - 2\lambda^2 - 3) = 0$
$(-\lambda^2-1) (\lambda^2-3) (\lambda^2+1) = 0$
Since $\lambda$ is real,$\lambda^2+1 \neq 0$. Thus,$\lambda^2-3 = 0$,which gives $\lambda^2 = 3$.
Therefore,$\lambda = \pm \sqrt{3}$.
So,$S = \{-\sqrt{3}, \sqrt{3}\}$.

(C) The line is given by $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$. The direction vector of the line is $\vec{b} = (1, 1, 2)$ and the normal to the plane $2x-4y+z=7$ is $\vec{n} = (2, -4, 1)$.
First,check if the line is parallel to the plane by calculating the dot product $\vec{b} \cdot \vec{n} = (1)(2) + (1)(-4) + (2)(1) = 2 - 4 + 2 = 0$. Since the dot product is $0$,the line is parallel to the plane.
For the line to lie in the plane,any point on the line must satisfy the plane equation. Let the point on the line be $(4, 2, k)$.
Substituting this point into the plane equation $2x-4y+z=7$:
$2(4) - 4(2) + k = 7$
$8 - 8 + k = 7$
$k = 7$.

(B) Let the direction ratios of the required line be $\langle a, b, c \rangle$.
Since the line is perpendicular to the lines with direction ratios $\langle 1, 2, 3 \rangle$ and $\langle -3, 2, 5 \rangle$,we have:
$1a + 2b + 3c = 0$
$-3a + 2b + 5c = 0$
Using the cross product to find the direction ratios $\langle a, b, c \rangle$:
$a = (2)(5) - (3)(2) = 10 - 6 = 4$
$b = (3)(-3) - (1)(5) = -9 - 5 = -14$
$c = (1)(2) - (2)(-3) = 2 + 6 = 8$
Thus,the direction ratios are $\langle 4, -14, 8 \rangle$,which simplifies to $\langle 2, -7, 4 \rangle$.
The equation of the line passing through $(2, 1, 3)$ with direction ratios $\langle 2, -7, 4 \rangle$ is:
$\frac{x-2}{2} = \frac{y-1}{-7} = \frac{z-3}{4}$
This can be rewritten as:
$\frac{x-2}{2} = \frac{1-y}{7} = \frac{z-3}{4}$

(C) We know that $\tan \left(\frac{7 \pi}{4}\right) = \tan \left(2 \pi - \frac{\pi}{4}\right) = -\tan \left(\frac{\pi}{4}\right) = -1$.
Therefore,$\cos ^{-1}\left(\tan \left(\frac{7 \pi}{4}\right)\right) = \cos ^{-1}(-1)$.
Since $\cos (\pi) = -1$,we have $\cos ^{-1}(-1) = \pi$.

(A) Let $\theta = \frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}$.
Then $2\theta = \cos ^{-1} \frac{\sqrt{5}}{3}$,which implies $\cos 2\theta = \frac{\sqrt{5}}{3}$.
We need to find $\tan \theta$.
Using the formula $\tan \theta = \sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}}$:
$\tan \theta = \sqrt{\frac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}} = \sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}$.
Rationalizing the denominator:
$\tan \theta = \sqrt{\frac{(3-\sqrt{5})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}} = \sqrt{\frac{(3-\sqrt{5})^2}{9-5}} = \sqrt{\frac{(3-\sqrt{5})^2}{4}} = \frac{3-\sqrt{5}}{2}$.

(A) Given equation: $\sin \left(\cot ^{-1}(x+1)\right)=\cos \left(\tan ^{-1} x\right)$
Let $\cot ^{-1}(x+1) = \theta$,then $\cot \theta = x+1$. Using the identity $\sin \theta = \frac{1}{\sqrt{1+\cot^2 \theta}}$,we get $\sin \theta = \frac{1}{\sqrt{1+(x+1)^2}} = \frac{1}{\sqrt{x^2+2x+2}}$.
Let $\tan ^{-1} x = \phi$,then $\tan \phi = x$. Using the identity $\cos \phi = \frac{1}{\sqrt{1+\tan^2 \phi}}$,we get $\cos \phi = \frac{1}{\sqrt{1+x^2}}$.
Equating both sides: $\frac{1}{\sqrt{x^2+2x+2}} = \frac{1}{\sqrt{x^2+1}}$.
Squaring both sides: $x^2+1 = x^2+2x+2$.
$2x = -1 \Rightarrow x = -\frac{1}{2}$.

(C) Given that $\vec{a} \perp (\vec{b}+\vec{c}) \implies \vec{a} \cdot (\vec{b}+\vec{c}) = 0 \implies \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0 \quad (i)$
$\vec{b} \perp (\vec{c}+\vec{a}) \implies \vec{b} \cdot (\vec{c}+\vec{a}) = 0 \implies \vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0 \quad (ii)$
$\vec{c} \perp (\vec{a}+\vec{b}) \implies \vec{c} \cdot (\vec{a}+\vec{b}) = 0 \implies \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0 \quad (iii)$
Adding $(i), (ii),$ and $(iii)$,we get:
$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \implies \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = 0 \quad (iv)$
Now,the length of $\vec{a}+\vec{b}+\vec{c}$ is given by:
$|\vec{a}+\vec{b}+\vec{c}| = \sqrt{|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})}$
Substituting the given values $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=5$ and the result from $(iv)$:
$|\vec{a}+\vec{b}+\vec{c}| = \sqrt{3^2+4^2+5^2 + 2(0)}$
$|\vec{a}+\vec{b}+\vec{c}| = \sqrt{9+16+25} = \sqrt{50} = 5\sqrt{2}$

(B) The projection of a vector $\vec{v} = p \hat{i} + q \hat{j} + r \hat{k}$ on the coordinate axes are the absolute values of its components.
The projection on the $x$-axis is $|p| = |4| = 4$.
The projection on the $y$-axis is $|q| = |-5| = 5$.
The projection on the $z$-axis is $|r| = |7| = 7$.
The sum of the lengths of these projections is $|p| + |q| + |r| = 4 + 5 + 7 = 16 \text{ units}$.

(C) Given the equation $2 \vec{a} + 3 \vec{b} - 5 \vec{c} = \vec{0}$.
Rearranging the terms,we get $2 \vec{a} + 3 \vec{b} = 5 \vec{c}$.
Dividing by $5$,we have $\vec{c} = \frac{2 \vec{a} + 3 \vec{b}}{5} = \frac{2 \vec{a} + 3 \vec{b}}{2 + 3}$.
This is in the form of the section formula for internal division,$\vec{r} = \frac{m \vec{b} + n \vec{a}}{m + n}$,where the point divides the segment joining $\vec{a}$ and $\vec{b}$ in the ratio $m:n$.
Here,$m = 3$ and $n = 2$.
Therefore,point $C$ divides the segment $AB$ in the ratio $3:2$ internally.

(A) First,find the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -4 \\ 6 & 5 & -2 \end{vmatrix} = \hat{i}(-2 - (-20)) - \hat{j}(-6 - (-24)) + \hat{k}(15 - 6) = 18 \hat{i} - 18 \hat{j} + 9 \hat{k}$
Next,find the magnitude of the cross product:
$|\vec{a} \times \vec{b}| = \sqrt{18^2 + (-18)^2 + 9^2} = \sqrt{324 + 324 + 81} = \sqrt{729} = 27$
The unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ is $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{18 \hat{i} - 18 \hat{j} + 9 \hat{k}}{27} = \frac{2}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{1}{3} \hat{k}$
The required vector with magnitude $3$ is $\pm 3 \left( \frac{2}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{1}{3} \hat{k} \right) = \pm(2 \hat{i} - 2 \hat{j} + \hat{k})$.

(B) Since $\vec{c}$ is a linear combination of $\vec{a}$ and $\vec{b}$,we can write $\vec{c} = m\vec{a} + n\vec{b}$ for some scalars $m$ and $n$.
Alternatively,since $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$,the scalar triple product $[\vec{a}, \vec{b}, \vec{c}] = 0$.
Calculating the determinant:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{vmatrix} = 0$
Expanding along the first row:
$1(1 - 2(x-2)) - 1(-1 - 2x) + 1((x-2) - (-x)) = 0$
$1(1 - 2x + 4) - 1(-1 - 2x) + 1(x - 2 + x) = 0$
$(5 - 2x) + (1 + 2x) + (2x - 2) = 0$
$2x + 4 = 0$
$2x = -4$
$x = -2$

(D) Let the position vectors of vertices $A, B, C$ be $\vec{a} = 4\hat{i} - 2\hat{j}$,$\vec{b} = \hat{i} + 4\hat{j} - 3\hat{k}$,and $\vec{c} = -\hat{i} + 5\hat{j} + \hat{k}$.
The angle $\angle ABC$ is the angle between vectors $\vec{BA}$ and $\vec{BC}$.
$\vec{BA} = \vec{a} - \vec{b} = (4-1)\hat{i} + (-2-4)\hat{j} + (0-(-3))\hat{k} = 3\hat{i} - 6\hat{j} + 3\hat{k}$.
$\vec{BC} = \vec{c} - \vec{b} = (-1-1)\hat{i} + (5-4)\hat{j} + (1-(-3))\hat{k} = -2\hat{i} + \hat{j} + 4\hat{k}$.
The cosine of the angle $\theta$ is given by $\cos \theta = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}| |\vec{BC}|}$.
$\vec{BA} \cdot \vec{BC} = (3)(-2) + (-6)(1) + (3)(4) = -6 - 6 + 12 = 0$.
Since the dot product is $0$,the vectors are perpendicular.
Therefore,$\theta = \cos^{-1}(0) = \frac{\pi}{2}$.

(C) Let the coordinates of point $B$ be $(x, y, z)$.
Given the position vector of point $A$ is $\vec{a} = (1\hat{i} + 2\hat{j} - 1\hat{k})$.
The vector $\overrightarrow{AB}$ is given by $\vec{b} - \vec{a}$.
So,$\overrightarrow{AB} = (x - 1)\hat{i} + (y - 2)\hat{j} + (z - (-1))\hat{k} = (x - 1)\hat{i} + (y - 2)\hat{j} + (z + 1)\hat{k}$.
We are given $\overrightarrow{AB} = 2\hat{i} + 3\hat{j} - 1\hat{k}$.
Comparing the components,we get:
$x - 1 = 2 \Rightarrow x = 3$
$y - 2 = 3 \Rightarrow y = 5$
$z + 1 = -1 \Rightarrow z = -2$
Therefore,the coordinates of $B$ are $(3, 5, -2)$.

(B) Given that $|\vec{u}+\vec{v}+\vec{w}|=0$.
Squaring both sides,we get $|\vec{u}+\vec{v}+\vec{w}|^2 = 0^2$.
Using the identity $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$,we have:
$|\vec{u}|^2+|\vec{v}|^2+|\vec{w}|^2+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}) = 0$.
Substituting the given values $|\vec{u}|=3, |\vec{v}|=4, |\vec{w}|=5$:
$3^2+4^2+5^2+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}) = 0$.
$9+16+25+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}) = 0$.
$50+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}) = 0$.
$2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}) = -50$.
$\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u} = -25$.
Taking the absolute value as requested:
$|\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}| = |-25| = 25$.

(B) Four points $A, B, C, D$ are coplanar if the scalar triple product of vectors $\vec{AB}, \vec{AC},$ and $\vec{AD}$ is zero,i.e.,$[\vec{AB}, \vec{AC}, \vec{AD}] = 0$.
Given points are $A(3,2,1), B(4, x, 5), C(4,2,-2), D(6,5,-1)$.
Vectors are:
$\vec{AB} = (4-3)\hat{i} + (x-2)\hat{j} + (5-1)\hat{k} = \hat{i} + (x-2)\hat{j} + 4\hat{k}$
$\vec{AC} = (4-3)\hat{i} + (2-2)\hat{j} + (-2-1)\hat{k} = \hat{i} + 0\hat{j} - 3\hat{k}$
$\vec{AD} = (6-3)\hat{i} + (5-2)\hat{j} + (-1-1)\hat{k} = 3\hat{i} + 3\hat{j} - 2\hat{k}$
For coplanarity,the determinant of these vectors must be zero:
$\begin{vmatrix} 1 & x-2 & 4 \\ 1 & 0 & -3 \\ 3 & 3 & -2 \end{vmatrix} = 0$
Expanding along the first row:
$1(0 - (-9)) - (x-2)(-2 - (-9)) + 4(3 - 0) = 0$
$1(9) - (x-2)(7) + 4(3) = 0$
$9 - 7x + 14 + 12 = 0$
$-7x + 35 = 0$
$7x = 35$
$x = 5$

(B) We are given the expression $(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$.
Using the distributive property of the dot product,we expand the expression:
$= (a \vec{b}) \cdot (a \vec{b}) - (a \vec{b}) \cdot (b \vec{a}) + (b \vec{a}) \cdot (a \vec{b}) - (b \vec{a}) \cdot (b \vec{a})$
$= a^2 (\vec{b} \cdot \vec{b}) - ab (\vec{b} \cdot \vec{a}) + ab (\vec{a} \cdot \vec{b}) - b^2 (\vec{a} \cdot \vec{a})$
Since the dot product is commutative,$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$,so the middle terms cancel out:
$= a^2 |\vec{b}|^2 - b^2 |\vec{a}|^2$
Wait,let us re-evaluate the expression given in the prompt: $(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$.
$= a^2 (\vec{b} \cdot \vec{b}) - b^2 (\vec{a} \cdot \vec{a})$
$= a^2 |\vec{b}|^2 - b^2 |\vec{a}|^2$
If $a$ and $b$ are scalars representing the magnitudes $|\vec{a}|$ and $|\vec{b}|$,then $a^2 |\vec{b}|^2 - b^2 |\vec{a}|^2 = |\vec{a}|^2 |\vec{b}|^2 - |\vec{b}|^2 |\vec{a}|^2 = 0$.

(B) We know that $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$.
Substituting the given values: $35 = \sqrt{26} \times 7 \times \sin \theta$.
$\sin \theta = \frac{35}{7 \sqrt{26}} = \frac{5}{\sqrt{26}}$.
Since $\cos^2 \theta = 1 - \sin^2 \theta$,we have $\cos^2 \theta = 1 - \frac{25}{26} = \frac{1}{26}$.
Thus,$\cos \theta = \frac{1}{\sqrt{26}}$ (assuming $\theta$ is acute).
Now,the dot product is given by $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$.
$\vec{a} \cdot \vec{b} = \sqrt{26} \times 7 \times \frac{1}{\sqrt{26}} = 7$.

(C) Given $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=\hat{i}+\hat{j}$,and $|\vec{c}-\vec{a}|=3$.
First,calculate $\vec{p} = \vec{a} \times \vec{b}$:
$\vec{p} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i} - 2\hat{j} + \hat{k}$.
The magnitude $|\vec{p}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = 3$.
Given $|\vec{p} \times \vec{c}| = 3$ and the angle between $\vec{p}$ and $\vec{c}$ is $\frac{\pi}{6}$,we have:
$|\vec{p}||\vec{c}| \sin(\frac{\pi}{6}) = 3$
$3 \cdot |\vec{c}| \cdot \frac{1}{2} = 3 \implies |\vec{c}| = 2$.
Now,use the condition $|\vec{c}-\vec{a}|=3$:
$|\vec{c}-\vec{a}|^2 = 3^2 = 9$
$|\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{a} \cdot \vec{c}) = 9$.
We know $|\vec{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = 3$,so $|\vec{a}|^2 = 9$.
Substituting the values: $4 + 9 - 2(\vec{a} \cdot \vec{c}) = 9$
$13 - 2(\vec{a} \cdot \vec{c}) = 9$
$2(\vec{a} \cdot \vec{c}) = 4$
$\vec{a} \cdot \vec{c} = 2$.

(B) We know that the magnitude of the difference of two vectors is given by the formula:
$|\bar{a}-\bar{b}| = \sqrt{|\bar{a}|^2 + |\bar{b}|^2 - 2(\bar{a} \cdot \bar{b})}$
Given that $|\bar{a}| = 3$,$|\bar{b}| = 2$,and $\bar{a} \cdot \bar{b} = 5$.
Substituting these values into the formula:
$|\bar{a}-\bar{b}| = \sqrt{3^2 + 2^2 - 2(5)}$
$|\bar{a}-\bar{b}| = \sqrt{9 + 4 - 10}$
$|\bar{a}-\bar{b}| = \sqrt{13 - 10}$
$|\bar{a}-\bar{b}| = \sqrt{3}$

(C) The angle between two vectors $\vec{a}$ and $\vec{b}$ is obtuse if and only if their dot product is negative,i.e.,$\vec{a} \cdot \vec{b} < 0$.
Given $\vec{a} = 2\lambda^2 \hat{i} + 4\lambda \hat{j} + \hat{k}$ and $\vec{b} = 7\hat{i} - 2\hat{j} + \lambda \hat{k}$.
Calculating the dot product:
$\vec{a} \cdot \vec{b} = (2\lambda^2)(7) + (4\lambda)(-2) + (1)(\lambda) < 0$
$\Rightarrow 14\lambda^2 - 8\lambda + \lambda < 0$
$\Rightarrow 14\lambda^2 - 7\lambda < 0$
$\Rightarrow 7\lambda(2\lambda - 1) < 0$
To solve this inequality,we find the critical points $\lambda = 0$ and $\lambda = \frac{1}{2}$.
Testing the intervals,the expression $7\lambda(2\lambda - 1)$ is negative for $\lambda \in \left(0, \frac{1}{2}\right)$.
Thus,the values of $\lambda$ lie in $\left(0, \frac{1}{2}\right)$.

(A) Given that $\bar{a}, \bar{b}, \bar{c}$ are unit vectors,so $|\bar{a}| = |\bar{b}| = |\bar{c}| = 1$.
We are given $|\bar{a}+\bar{b}+\bar{c}|=1$. Squaring both sides,we get:
$|\bar{a}+\bar{b}+\bar{c}|^2 = 1^2$
$|\bar{a}|^2+|\bar{b}|^2+|\bar{c}|^2+2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = 1$
Since $\bar{b} \perp \bar{c}$,we have $\bar{b} \cdot \bar{c} = 0$.
Also,$\bar{a} \cdot \bar{b} = |\bar{a}||\bar{b}| \cos \alpha = \cos \alpha$ and $\bar{a} \cdot \bar{c} = |\bar{a}||\bar{c}| \cos \beta = \cos \beta$.
Substituting these values into the equation:
$1+1+1+2(\cos \alpha + 0 + \cos \beta) = 1$
$3 + 2(\cos \alpha + \cos \beta) = 1$
$2(\cos \alpha + \cos \beta) = 1 - 3$
$2(\cos \alpha + \cos \beta) = -2$
$\cos \alpha + \cos \beta = -1$

(B) Any vector $\vec{V}$ in the plane of $\vec{a}$ and $\vec{b}$ can be expressed as a linear combination: $\vec{V} = x\vec{a} + y\vec{b}$. For simplicity,we can write $\vec{V} = \vec{a} + \lambda\vec{b}$ (assuming the coefficient of $\vec{a}$ is non-zero).
$\vec{V} = (\hat{i}+\hat{j}+\hat{k}) + \lambda(\hat{i}-\hat{j}+\hat{k}) = (1+\lambda)\hat{i} + (1-\lambda)\hat{j} + (1+\lambda)\hat{k}$.
The projection of $\vec{V}$ on $\vec{c}$ is given by $\frac{\vec{V} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}}$.
Given $\vec{c} = \hat{i}-\hat{j}-\hat{k}$,we have $|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$.
So,$\frac{\vec{V} \cdot \vec{c}}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow \vec{V} \cdot \vec{c} = 1$.
$((1+\lambda)\hat{i} + (1-\lambda)\hat{j} + (1+\lambda)\hat{k}) \cdot (\hat{i}-\hat{j}-\hat{k}) = 1$.
$(1+\lambda) - (1-\lambda) - (1+\lambda) = 1$.
$1 + \lambda - 1 + \lambda - 1 - \lambda = 1$.
$\lambda - 1 = 1 \Rightarrow \lambda = 2$.
Substituting $\lambda = 2$ into the expression for $\vec{V}$:
$\vec{V} = (1+2)\hat{i} + (1-2)\hat{j} + (1+2)\hat{k} = 3\hat{i} - \hat{j} + 3\hat{k}$.

(A) Given $\vec{b} \times \vec{c} = \vec{b} \times \vec{a}$.
This implies $\vec{b} \times (\vec{c} - \vec{a}) = \vec{0}$.
Therefore,$\vec{c} - \vec{a} = \lambda \vec{b}$ for some scalar $\lambda$,so $\vec{c} = \vec{a} + \lambda \vec{b}$.
Given $\vec{c} \cdot \vec{a} = 0$,we have $(\vec{a} + \lambda \vec{b}) \cdot \vec{a} = 0$,which means $\vec{a} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$.
Calculating the dot products:
$\vec{a} \cdot \vec{a} = (1)^2 + (-2)^2 + (1)^2 = 1 + 4 + 1 = 6$.
$\vec{b} \cdot \vec{a} = (1)(1) + (-1)(-2) + (1)(1) = 1 + 2 + 1 = 4$.
Substituting these values: $6 + \lambda(4) = 0 \Rightarrow 4\lambda = -6 \Rightarrow \lambda = -\frac{3}{2}$.
Now,$\vec{c} \cdot \vec{b} = (\vec{a} + \lambda \vec{b}) \cdot \vec{b} = \vec{a} \cdot \vec{b} + \lambda (\vec{b} \cdot \vec{b})$.
$\vec{b} \cdot \vec{b} = (1)^2 + (-1)^2 + (1)^2 = 3$.
$\vec{c} \cdot \vec{b} = 4 + (-\frac{3}{2})(3) = 4 - \frac{9}{2} = -\frac{1}{2}$.

(B) Since $\vec{r}$ lies in the plane of $\vec{a}$ and $\vec{b}$,we can write $\vec{r} = t\vec{a} + u\vec{b}$ for some scalars $t$ and $u$.
$\vec{r} = t(\hat{i}+\hat{j}+\hat{k}) + u(\hat{i}-\hat{j}+\hat{k}) = (t+u)\hat{i} + (t-u)\hat{j} + (t+u)\hat{k} \dots (i)$
Given that the projection of $\vec{r}$ on $\vec{c}$ is $\frac{1}{\sqrt{3}}$,we have $\frac{\vec{r} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}}$.
$|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$.
$\vec{r} \cdot \vec{c} = (t+u)(1) + (t-u)(-1) + (t+u)(-1) = t+u - t+u - t-u = u-t$.
So,$\frac{u-t}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow u-t = 1 \Rightarrow u = t+1$.
Substituting $u = t+1$ into equation $(i)$:
$\vec{r} = (t + t + 1)\hat{i} + (t - (t + 1))\hat{j} + (t + t + 1)\hat{k}$
$\vec{r} = (2t+1)\hat{i} - \hat{j} + (2t+1)\hat{k}$.

(B) Given $\vec{a}+\vec{b} = \lambda \vec{c} \quad \dots(i)$ and $\vec{b}+\vec{c} = \mu \vec{a} \quad \dots(ii)$.
Subtracting $(ii)$ from $(i)$,we get $\vec{a}-\vec{c} = \lambda \vec{c} - \mu \vec{a}$.
Rearranging terms,$(1+\mu)\vec{a} = (1+\lambda)\vec{c}$.
Since two vectors are collinear,let $\vec{a}$ and $\vec{b}$ be collinear,then $\vec{b} = k\vec{a}$.
Given $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{2}$,we have $|k\vec{a}| = |\vec{a}| \Rightarrow |k|=1$,so $k=1$ or $k=-1$.
If $k=1$,$\vec{b}=\vec{a}$,then $\vec{a}+\vec{a} = 2\vec{a}$ is collinear with $\vec{c}$,so $\vec{c} = \pm \vec{a}$.
If $\vec{c} = -\vec{a}$,then $\vec{a}+\vec{b}+\vec{c} = \vec{a}+\vec{a}-\vec{a} = \vec{a} \neq 0$. However,the condition $\vec{a}+\vec{b}+\vec{c}=0$ is derived from the collinearity constraints.
Using $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$.
Since $\vec{a}+\vec{b}+\vec{c}=0$,we have $0 = 2+2+2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$.
Thus,$2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = -6$.
Therefore,$\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} = -3$.

(D) The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula: $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
Given that $|\vec{a}| = \sqrt{2}$,$|\vec{b}| = \sqrt{2}$,and $\vec{a} \cdot \vec{b} = -1$.
Substituting these values into the formula:
$\cos \theta = \frac{-1}{\sqrt{2} \times \sqrt{2}} = \frac{-1}{2}$.
Since $\cos \theta = -\frac{1}{2}$,the angle $\theta$ is $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$.

(C) The magnitude of the cross product of two vectors $\vec{u}$ and $\vec{v}$ is given by the formula:
$|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin \theta$
From the given figure,we have:
$|\vec{u}| = 4$
$|\vec{v}| = 5$
$\theta = 150^{\circ}$
Substituting these values into the formula:
$|\vec{u} \times \vec{v}| = 4 \times 5 \times \sin 150^{\circ}$
Since $\sin 150^{\circ} = \sin(180^{\circ} - 30^{\circ}) = \sin 30^{\circ} = \frac{1}{2}$,we get:
$|\vec{u} \times \vec{v}| = 20 \times \frac{1}{2} = 10$

(C) Given: $|\vec{a}| = \sqrt{26}$,$|\vec{b}| = 7$,and $|\vec{a} \times \vec{b}| = 35$.
We know that $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$.
Substituting the values: $35 = \sqrt{26} \times 7 \times \sin \theta$.
$\sin \theta = \frac{35}{7 \sqrt{26}} = \frac{5}{\sqrt{26}}$.
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$:
$\cos^2 \theta = 1 - (\frac{5}{\sqrt{26}})^2 = 1 - \frac{25}{26} = \frac{1}{26}$.
Therefore,$\cos \theta = \pm \frac{1}{\sqrt{26}}$.
Now,the dot product is given by $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$.
$\vec{a} \cdot \vec{b} = \sqrt{26} \times 7 \times (\pm \frac{1}{\sqrt{26}}) = \pm 7$.

(C) Given that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.
Rearranging the terms,we get $\vec{a}+\vec{b}=-\vec{c}$.
Squaring both sides,we have $|\vec{a}+\vec{b}|^2 = |-\vec{c}|^2$,which implies $|\vec{a}|^2+|\vec{b}|^2+2(\vec{a} \cdot \vec{b}) = |\vec{c}|^2$.
Substituting the given magnitudes $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$:
$3^2+5^2+2|\vec{a}||\vec{b}| \cos \theta = 7^2$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
$9+25+2(3)(5) \cos \theta = 49$.
$34+30 \cos \theta = 49$.
$30 \cos \theta = 49-34 = 15$.
$\cos \theta = \frac{15}{30} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \frac{\pi}{3}$.

(D) Given: $\vec{a}+2 \vec{b}+3 \vec{c}=\vec{0} \quad \dots(1)$
Taking the cross product with $\vec{b}$ on both sides:
$\vec{a} \times \vec{b} + 2(\vec{b} \times \vec{b}) + 3(\vec{c} \times \vec{b}) = \vec{0} \times \vec{b}$
$\Rightarrow \vec{a} \times \vec{b} + 0 - 3(\vec{b} \times \vec{c}) = \vec{0}$
$\Rightarrow \vec{a} \times \vec{b} = 3(\vec{b} \times \vec{c}) \quad \dots(2)$
Taking the cross product with $\vec{c}$ on both sides of $(1)$:
$\vec{a} \times \vec{c} + 2(\vec{b} \times \vec{c}) + 3(\vec{c} \times \vec{c}) = \vec{0} \times \vec{c}$
$\Rightarrow \vec{a} \times \vec{c} + 2(\vec{b} \times \vec{c}) + 0 = \vec{0}$
$\Rightarrow \vec{c} \times \vec{a} = 2(\vec{b} \times \vec{c}) \quad \dots(3)$
Now,substitute $(2)$ and $(3)$ into the given expression $(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})$:
$= 3(\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) + 2(\vec{b} \times \vec{c})$
$= (3+1+2)(\vec{b} \times \vec{c}) = 6(\vec{b} \times \vec{c})$
Comparing this with $\lambda(\vec{b} \times \vec{c})$,we get $\lambda = 6$.

(A) Let the vertices be $A = (1, y, 0)$,$B = (1, 0, 2)$,and $C = (0, 3, 1)$.
$\overrightarrow{AB} = (1-1)\hat{i} + (0-y)\hat{j} + (2-0)\hat{k} = -y\hat{j} + 2\hat{k}$.
$\overrightarrow{AC} = (0-1)\hat{i} + (3-y)\hat{j} + (1-0)\hat{k} = -\hat{i} + (3-y)\hat{j} + \hat{k}$.
The area of the triangle is given by $\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{6}$.
$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -y & 2 \\ -1 & 3-y & 1 \end{vmatrix} = \hat{i}(-y - 2(3-y)) - \hat{j}(0 - (-2)) + \hat{k}(0 - y) = \hat{i}(y-6) - 2\hat{j} - y\hat{k}$.
$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(y-6)^2 + (-2)^2 + (-y)^2} = \sqrt{y^2 - 12y + 36 + 4 + y^2} = \sqrt{2y^2 - 12y + 40}$.
Since $\frac{1}{2} \sqrt{2y^2 - 12y + 40} = \sqrt{6}$,we have $\sqrt{2y^2 - 12y + 40} = 2\sqrt{6} = \sqrt{24}$.
Squaring both sides: $2y^2 - 12y + 40 = 24 \Rightarrow 2y^2 - 12y + 16 = 0 \Rightarrow y^2 - 6y + 8 = 0$.
Factoring: $(y-2)(y-4) = 0$,so $y = 2, 4$.

(C) Given $|(\vec{a} \times \vec{b}) \cdot \vec{c}| = |\vec{a}||\vec{b}||\vec{c}|$.
Since $|(\vec{a} \times \vec{b}) \cdot \vec{c}| = |\vec{a} \times \vec{b}| |\vec{c}| |\cos \theta|$,where $\theta$ is the angle between $(\vec{a} \times \vec{b})$ and $\vec{c}$,we have $|\vec{a} \times \vec{b}| |\vec{c}| |\cos \theta| = |\vec{a}||\vec{b}||\vec{c}|$.
Substituting $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \phi$,where $\phi$ is the angle between $\vec{a}$ and $\vec{b}$,we get $|\vec{a}||\vec{b}| \sin \phi |\vec{c}| |\cos \theta| = |\vec{a}||\vec{b}||\vec{c}|$.
This implies $\sin \phi = 1$ and $|\cos \theta| = 1$.
Thus,$\phi = 90^{\circ}$ (meaning $\vec{a} \perp \vec{b}$) and $\theta = 0^{\circ}$ or $180^{\circ}$ (meaning $\vec{c}$ is parallel to the normal of the plane containing $\vec{a}$ and $\vec{b}$).
Since $\vec{c}$ is parallel to $\vec{a} \times \vec{b}$,it follows that $\vec{c} \perp \vec{a}$ and $\vec{c} \perp \vec{b}$.
Therefore,$\vec{a} \cdot \vec{b} = 0$,$\vec{b} \cdot \vec{c} = 0$,and $\vec{c} \cdot \vec{a} = 0$.

(A) Given $\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}$.
We know that $\vec{u} \times \vec{v} = -\vec{v} \times \vec{u}$,so $(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b} = -\vec{b} \times (2 \hat{i}+3 \hat{j}+4 \hat{k})$.
Substituting this into the equation,we get $\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k}) + \vec{b} \times(2 \hat{i}+3 \hat{j}+4 \hat{k}) = \vec{0}$.
This implies $(\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \hat{k}) = \vec{0}$.
This means the vector $(\vec{a}+\vec{b})$ is parallel to $(2 \hat{i}+3 \hat{j}+4 \hat{k})$.
Let $\vec{a}+\vec{b} = \lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ for some scalar $\lambda$.
Taking the magnitude on both sides,$|\vec{a}+\vec{b}| = |\lambda| \sqrt{2^2+3^2+4^2} = |\lambda| \sqrt{4+9+16} = |\lambda| \sqrt{29}$.
Given $|\vec{a}+\vec{b}| = \sqrt{29}$,we have $|\lambda| \sqrt{29} = \sqrt{29}$,which gives $\lambda = \pm 1$.
Thus,$\vec{a}+\vec{b} = \pm(2 \hat{i}+3 \hat{j}+4 \hat{k})$.
Now,calculate the dot product: $(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) = \pm(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$.
$= \pm((2)(-7) + (3)(2) + (4)(3)) = \pm(-14 + 6 + 12) = \pm 4$.
Therefore,a possible value is $4$.

(C) Given that $(\vec{a}+\lambda \vec{b})$ is perpendicular to $\vec{c}$,their dot product must be zero:
$(\vec{a}+\lambda \vec{b}) \cdot \vec{c} = 0$
First,calculate $(\vec{a}+\lambda \vec{b})$:
$\vec{a}+\lambda \vec{b} = (2 \hat{i}+2 \hat{j}+3 \hat{k}) + \lambda(-\hat{i}+2 \hat{j}+\hat{k}) = (2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}$
Now,take the dot product with $\vec{c} = 3 \hat{i} + \hat{j} + 0 \hat{k}$:
$((2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}) \cdot (3 \hat{i} + \hat{j} + 0 \hat{k}) = 0$
$3(2-\lambda) + 1(2+2\lambda) + 0(3+\lambda) = 0$
$6 - 3\lambda + 2 + 2\lambda = 0$
$8 - \lambda = 0$
$\lambda = 8$

(D) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
Also,$\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
Squaring both sides of the equation,we get:
$|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{0}|^2$
$(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$
$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
Substituting the values $|\vec{a}| = 1, |\vec{b}| = 1, |\vec{c}| = 1$:
$1^2 + 1^2 + 1^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
$3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3$
$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$

(C) Given the condition: $\overrightarrow{OP} \cdot \overrightarrow{OQ} + \overrightarrow{OR} \cdot \overrightarrow{OS} = \overrightarrow{OR} \cdot \overrightarrow{OP} + \overrightarrow{OQ} \cdot \overrightarrow{OS} = \overrightarrow{OQ} \cdot \overrightarrow{OR} + \overrightarrow{OP} \cdot \overrightarrow{OS}$.
Consider the first equality: $\overrightarrow{OP} \cdot \overrightarrow{OQ} + \overrightarrow{OR} \cdot \overrightarrow{OS} = \overrightarrow{OR} \cdot \overrightarrow{OP} + \overrightarrow{OQ} \cdot \overrightarrow{OS}$.
Rearranging the terms: $\overrightarrow{OP} \cdot \overrightarrow{OQ} - \overrightarrow{OR} \cdot \overrightarrow{OP} = \overrightarrow{OQ} \cdot \overrightarrow{OS} - \overrightarrow{OR} \cdot \overrightarrow{OS}$.
Factoring out the common vectors: $\overrightarrow{OP} \cdot (\overrightarrow{OQ} - \overrightarrow{OR}) = \overrightarrow{OS} \cdot (\overrightarrow{OQ} - \overrightarrow{OR})$.
This implies: $(\overrightarrow{OP} - \overrightarrow{OS}) \cdot (\overrightarrow{OQ} - \overrightarrow{OR}) = 0$.
Since $\overrightarrow{OP} - \overrightarrow{OS} = \overrightarrow{SP}$ and $\overrightarrow{OQ} - \overrightarrow{OR} = \overrightarrow{RQ}$,we have $\overrightarrow{SP} \cdot \overrightarrow{RQ} = 0$,which means $\overrightarrow{PS} \perp \overrightarrow{QR}$.
Similarly,by considering the other parts of the equality,we get $\overrightarrow{QS} \perp \overrightarrow{PR}$ and $\overrightarrow{RS} \perp \overrightarrow{PQ}$.
Since $S$ is the point where the altitudes of the triangle $PQR$ intersect,$S$ is the orthocentre.

(D) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by $|\vec{a} \times \vec{b}| = 15$.
Now,consider the parallelogram with adjacent sides $(3 \vec{a} + \vec{b})$ and $(\vec{a} + 3 \vec{b})$.
The area of this parallelogram is $|(3 \vec{a} + \vec{b}) \times (\vec{a} + 3 \vec{b})|$.
Expanding the cross product:
$|(3 \vec{a} + \vec{b}) \times (\vec{a} + 3 \vec{b})| = |3 \vec{a} \times \vec{a} + 9 \vec{a} \times \vec{b} + \vec{b} \times \vec{a} + 3 \vec{b} \times \vec{b}|$.
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:
$|0 + 9(\vec{a} \times \vec{b}) - (\vec{a} \times \vec{b}) + 0| = |8(\vec{a} \times \vec{b})| = 8 |\vec{a} \times \vec{b}|$.
Substituting the given area $|\vec{a} \times \vec{b}| = 15$:
Area $= 8 \times 15 = 120$ sq. units.

(A) Let the direction angles of the unit vector $\hat{a}$ be $\alpha, \beta, \text{ and } \gamma$.
Given $\alpha = \frac{\pi}{3}$,$\beta = \frac{\pi}{4}$,and $\gamma = \theta$.
We know that for any unit vector,the sum of the squares of the direction cosines is $1$:
$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$
Substituting the given values:
$\cos^2 \left(\frac{\pi}{3}\right) + \cos^2 \left(\frac{\pi}{4}\right) + \cos^2 \theta = 1$
$\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \theta = 1$
$\frac{1}{4} + \frac{1}{2} + \cos^2 \theta = 1$
$\frac{3}{4} + \cos^2 \theta = 1$
$\cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4}$
$\cos \theta = \pm \frac{1}{2}$
Since $\theta \in (0, \pi)$,we have two possible values:
If $\cos \theta = \frac{1}{2}$,then $\theta = \frac{\pi}{3}$.
If $\cos \theta = -\frac{1}{2}$,then $\theta = \frac{2\pi}{3}$.
Comparing with the given options,$\frac{2\pi}{3}$ is the correct value.

(D) Using the vector triple product identity $(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$,let $\vec{u} = \vec{a}$,$\vec{v} = \vec{b}$,and $\vec{w} = (\vec{a} \times \vec{c})$.
Then $(\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c}) = [\vec{a} \vec{b} (\vec{a} \times \vec{c})] \vec{a} - [\vec{b} (\vec{a} \times \vec{c}) \vec{a}] \vec{a} = [\vec{a} \vec{b} \vec{c}] \vec{a} - 0 = [\vec{a} \vec{b} \vec{c}] \vec{a}$.
Now,$(\vec{a}-\vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c})] = (\vec{a}-\vec{b}) \cdot ((\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}) = [\vec{a} \vec{b} \vec{c}] (\vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{a})$.
Calculating the scalar triple product $[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 1 & 1 \\ -1 & 2 & -2 \\ 2 & -1 & 2 \end{vmatrix} = 1(4-2) - 1(-2+4) + 1(1-4) = 2 - 2 - 3 = -3$.
Calculating $(\vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{a}) = (3) - (-1+2-2) = 3 - (-1) = 4$.
Thus,the result is $(-3) \times 4 = -12$.

(C) Given equations are:
$(i)$ $\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$
(ii) $\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$
From (ii),we know that $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$,so $\vec{r} \times \vec{b} = -(\vec{b} \times \vec{a})$.
Adding $(i)$ and (ii):
$(\vec{r} \times \vec{a}) + (\vec{r} \times \vec{b}) = (\vec{b} \times \vec{a}) - (\vec{b} \times \vec{a}) = \vec{0}$
$\vec{r} \times (\vec{a} + \vec{b}) = \vec{0}$
This implies $\vec{r}$ is parallel to $(\vec{a} + \vec{b})$.
Given $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = 2\hat{i} - \hat{k}$,then $\vec{a} + \vec{b} = (1+2)\hat{i} + 1\hat{j} - 1\hat{k} = 3\hat{i} + \hat{j} - \hat{k}$.
Thus,$\vec{r} = \lambda(3\hat{i} + \hat{j} - \hat{k})$.
For the intersection point,we test the options. For $\lambda = 1$,$\vec{r} = 3\hat{i} + \hat{j} - \hat{k}$,which corresponds to the point $(3, 1, -1)$.
Checking this in equation $(i)$: $(3\hat{i} + \hat{j} - \hat{k}) \times (\hat{i} + \hat{j}) = 3(\hat{i} \times \hat{j}) + 1(\hat{j} \times \hat{i}) - 1(\hat{k} \times \hat{i}) - 1(\hat{k} \times \hat{j}) = 3\hat{k} - \hat{k} - \hat{j} + \hat{i} = \hat{i} - \hat{j} + 2\hat{k}$.
And $\vec{b} \times \vec{a} = (2\hat{i} - \hat{k}) \times (\hat{i} + \hat{j}) = 2(\hat{i} \times \hat{j}) - (\hat{k} \times \hat{i}) - (\hat{k} \times \hat{j}) = 2\hat{k} - \hat{j} + \hat{i} = \hat{i} - \hat{j} + 2\hat{k}$.
Since both sides are equal,the point $(3, 1, -1)$ is the intersection point.

(A) We are given $|\vec{a}|=5$,$|\vec{b}|=13$,and $|\vec{a} \times \vec{b}|=25$.
Using the formula $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$,we have:
$25 = 5 \times 13 \sin \theta$
$25 = 65 \sin \theta$
$\sin \theta = \frac{25}{65} = \frac{5}{13}$.
Since $\frac{\pi}{2} < \theta < \pi$,the angle $\theta$ lies in the second quadrant,where $\cos \theta$ is negative.
Using $\cos^2 \theta = 1 - \sin^2 \theta$:
$\cos \theta = -\sqrt{1 - (\frac{5}{13})^2} = -\sqrt{1 - \frac{25}{169}} = -\sqrt{\frac{144}{169}} = -\frac{12}{13}$.
Now,the dot product is given by $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$:
$\vec{a} \cdot \vec{b} = 5 \times 13 \times (-\frac{12}{13}) = -60$.

(A) First,we find the vectors $\overline{PQ}$ and $\overline{AB}$:
$\overline{PQ} = (3 - (-2)) \hat{i} + (2 - 1) \hat{j} + (5 - 3) \hat{k} = 5 \hat{i} + \hat{j} + 2 \hat{k}$
$\overline{AB} = (7 - 4) \hat{i} + (-5 - (-3)) \hat{j} + (-1 - 5) \hat{k} = 3 \hat{i} - 2 \hat{j} - 6 \hat{k}$
The vector projection of $\overline{PQ}$ on $\overline{AB}$ is given by the formula:
$\text{Vector Projection} = \frac{(\overline{PQ} \cdot \overline{AB}) \overline{AB}}{|\overline{AB}|^2}$
Calculate the dot product $\overline{PQ} \cdot \overline{AB}$:
$\overline{PQ} \cdot \overline{AB} = (5)(3) + (1)(-2) + (2)(-6) = 15 - 2 - 12 = 1$
Calculate the magnitude squared $|\overline{AB}|^2$:
$|\overline{AB}|^2 = 3^2 + (-2)^2 + (-6)^2 = 9 + 4 + 36 = 49$
Thus,the vector projection is:
$\frac{1}{49}(3 \hat{i} - 2 \hat{j} - 6 \hat{k})$

(C) Let $\vec{a} = 2 \hat{i}-3 \hat{j}+6 \hat{k}$.
Since $\vec{b}$ is collinear to $\vec{a}$,we can write $\vec{b} = \lambda \vec{a}$ for some scalar $\lambda$.
The magnitude of $\vec{a}$ is $|\vec{a}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
The unit vector in the direction of $\vec{a}$ is $\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2 \hat{i}-3 \hat{j}+6 \hat{k}}{7}$.
Given $|\vec{b}| = 14$,the vector $\vec{b}$ can be expressed as $\vec{b} = \pm |\vec{b}| \hat{a}$.
Taking the positive direction,$\vec{b} = 14 \left( \frac{2 \hat{i}-3 \hat{j}+6 \hat{k}}{7} \right) = 2(2 \hat{i}-3 \hat{j}+6 \hat{k}) = 4 \hat{i}-6 \hat{j}+12 \hat{k}$.

(B) Let $I = \int_0^1 \frac{8 \log (1+x)}{1+x^2} \,dx$.
Substitute $x = \tan \theta$, so $dx = \sec^2 \theta \,d\theta$.
When $x = 0$, $\theta = 0$; when $x = 1$, $\theta = \frac{\pi}{4}$.
$I = \int_0^{\frac{\pi}{4}} \frac{8 \log (1+\tan \theta)}{1+\tan^2 \theta} \cdot \sec^2 \theta \,d\theta = \int_0^{\frac{\pi}{4}} 8 \log (1+\tan \theta) \,d\theta \quad \dots(i)$
Using the property $\int_0^a f(\theta) \,d\theta = \int_0^a f(a-\theta) \,d\theta$:
$I = \int_0^{\frac{\pi}{4}} 8 \log \left(1 + \tan\left(\frac{\pi}{4} - \theta\right)\right) \,d\theta$
Since $\tan\left(\frac{\pi}{4} - \theta\right) = \frac{1-\tan \theta}{1+\tan \theta}$, we have:
$I = \int_0^{\frac{\pi}{4}} 8 \log \left(1 + \frac{1-\tan \theta}{1+\tan \theta}\right) \,d\theta = \int_0^{\frac{\pi}{4}} 8 \log \left(\frac{2}{1+\tan \theta}\right) \,d\theta \quad \dots(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\frac{\pi}{4}} 8 \left[ \log(1+\tan \theta) + \log\left(\frac{2}{1+\tan \theta}\right) \right] \,d\theta$
$2I = 8 \int_0^{\frac{\pi}{4}} \log \left( (1+\tan \theta) \cdot \frac{2}{1+\tan \theta} \right) \,d\theta$
$2I = 8 \int_0^{\frac{\pi}{4}} \log 2 \,d\theta = 8 \log 2 [\theta]_0^{\frac{\pi}{4}} = 8 \log 2 \cdot \frac{\pi}{4} = 2\pi \log 2$.
Therefore, $I = \pi \log 2$.

(C) The scalar triple product is defined as $[\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})$.
Since $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$,$\vec{a}$ must be parallel to the vector $\vec{b} \times \vec{c}$.
Thus,the angle between $\vec{a}$ and $\vec{b} \times \vec{c}$ is $0$ or $\pi$.
Assuming $\vec{a}$ is in the direction of $\vec{b} \times \vec{c}$,we have:
$[\vec{a} \vec{b} \vec{c}] = |\vec{a}| |\vec{b} \times \vec{c}| \cos(0) = |\vec{a}| |\vec{b}| |\vec{c}| \sin\left(\frac{5 \pi}{6}\right)$.
Substituting the given values:
$[\vec{a} \vec{b} \vec{c}] = 5 \times 3 \times 4 \times \sin\left(\frac{5 \pi}{6}\right)$.
Since $\sin\left(\frac{5 \pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.
$[\vec{a} \vec{b} \vec{c}] = 60 \times \frac{1}{2} = 30$.

(A) The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is given by the determinant of the components of the vectors $\vec{a}, \vec{b},$ and $\vec{c}$.
$[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{vmatrix}$
Applying the column operation $C_3 \to C_3 + C_1$:
$[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 0 & 0 \\ x & 1 & 1 \\ y & x & 1+x \end{vmatrix}$
Expanding along the first row:
$[\vec{a} \vec{b} \vec{c}] = 1 \cdot \begin{vmatrix} 1 & 1 \\ x & 1+x \end{vmatrix} - 0 + 0$
$[\vec{a} \vec{b} \vec{c}] = (1+x) - x = 1$
Since the result is $1$,which is a constant,the value of $[\vec{a} \vec{b} \vec{c}]$ depends on neither $x$ nor $y$.

(D) The volume of a tetrahedron with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,$C(x_3, y_3, z_3)$,and $D(x_4, y_4, z_4)$ is given by the formula:
$V = \frac{1}{6} |(\vec{AB}) \cdot (\vec{AC} \times \vec{AD})| = \frac{1}{6} \left| \det \begin{bmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \\ x_4-x_1 & y_4-y_1 & z_4-z_1 \end{bmatrix} \right|$
Given vertices are $A(1, -6, 10)$,$B(-1, -3, 7)$,$C(5, -1, \lambda)$,and $D(7, -4, 7)$.
Calculating the vectors:
$\vec{AB} = (-1-1, -3-(-6), 7-10) = (-2, 3, -3)$
$\vec{AC} = (5-1, -1-(-6), \lambda-10) = (4, 5, \lambda-10)$
$\vec{AD} = (7-1, -4-(-6), 7-10) = (6, 2, -3)$
The volume is $11$,so:
$11 = \frac{1}{6} \left| \det \begin{bmatrix} -2 & 3 & -3 \\ 4 & 5 & \lambda-10 \\ 6 & 2 & -3 \end{bmatrix} \right|$
$66 = | -2(-15 - 2(\lambda-10)) - 3(-12 - 6(\lambda-10)) - 3(8 - 30) |$
$66 = | -2(-15 - 2\lambda + 20) - 3(-12 - 6\lambda + 60) - 3(-22) |$
$66 = | -2(5 - 2\lambda) - 3(48 - 6\lambda) + 66 |$
$66 = | -10 + 4\lambda - 144 + 18\lambda + 66 |$
$66 = | 22\lambda - 88 |$
Dividing by $22$: $3 = | \lambda - 4 |$
This gives two cases:
$1) \lambda - 4 = 3 \Rightarrow \lambda = 7$
$2) \lambda - 4 = -3 \Rightarrow \lambda = 1$
Since the options provided include $7$,we select $\lambda = 7$.

(A) The volume $V$ of a parallelepiped formed by vectors $\vec{u}, \vec{v}, \vec{w}$ is given by the absolute value of their scalar triple product: $V = |\vec{u} \cdot (\vec{v} \times \vec{w})|$.
The vectors are $\vec{u} = (1, a, 1)$,$\vec{v} = (0, 1, a)$,and $\vec{w} = (a, 0, 1)$.
The scalar triple product is given by the determinant:
$V(a) = \left|\begin{array}{ccc} 1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1 \end{array}\right|$
Expanding along the first row:
$V(a) = 1(1 - 0) - a(0 - a^2) + 1(0 - a) = 1 + a^3 - a = a^3 - a + 1$.
To find the minimum volume,we find the derivative $V'(a)$ and set it to $0$:
$V'(a) = 3a^2 - 1 = 0$.
$3a^2 = 1 \Rightarrow a^2 = \frac{1}{3} \Rightarrow a = \pm \frac{1}{\sqrt{3}}$.
Using the second derivative test: $V''(a) = 6a$.
For $a = \frac{1}{\sqrt{3}}$,$V''(a) = 6(\frac{1}{\sqrt{3}}) > 0$,which indicates a local minimum.
Therefore,the volume is minimum at $a = \frac{1}{\sqrt{3}}$.