MHT CET 2023 Mathematics Question Paper with Answer and Solution

(A) Let the given line be $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4} = \lambda$.
Any point on the line is given by $(x, y, z) = (2\lambda+1, -3\lambda+2, 4\lambda-5)$.
Since this point lies on the plane $2x+4y-z=3$,we substitute these coordinates into the plane equation:
$2(2\lambda+1) + 4(-3\lambda+2) - (4\lambda-5) = 3$.
Expanding the terms:
$4\lambda + 2 - 12\lambda + 8 - 4\lambda + 5 = 3$.
Combining the $\lambda$ terms and constants:
$-12\lambda + 15 = 3$.
$-12\lambda = 3 - 15$.
$-12\lambda = -12$.
$\lambda = 1$.
Substituting $\lambda = 1$ back into the coordinates:
$x = 2(1) + 1 = 3$.
$y = -3(1) + 2 = -1$.
$z = 4(1) - 5 = -1$.
Thus,the required coordinates are $(3, -1, -1)$.

(A) The line $L$ is the intersection of two planes: $2x + 3y + z = 1$ and $x + 3y + 2z = 2$.
The direction vector $\vec{v}$ of the line $L$ is given by the cross product of the normals to the planes,$\vec{n_1} = (2, 3, 1)$ and $\vec{n_2} = (1, 3, 2)$.
$\vec{v} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{vmatrix} = \hat{i}(6-3) - \hat{j}(4-1) + \hat{k}(6-3) = 3\hat{i} - 3\hat{j} + 3\hat{k}$.
We can simplify the direction vector to $\vec{v} = (1, -1, 1)$.
The line $L$ makes an angle $\alpha$ with the positive $X$-axis,which has the direction vector $\vec{u} = (1, 0, 0)$.
The cosine of the angle $\alpha$ is given by $\cos \alpha = \frac{|\vec{v} \cdot \vec{u}|}{|\vec{v}| |\vec{u}|}$.
$\cos \alpha = \frac{|(1)(1) + (-1)(0) + (1)(0)|}{\sqrt{1^2 + (-1)^2 + 1^2} \cdot \sqrt{1^2 + 0^2 + 0^2}} = \frac{1}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}}$.
Therefore,$\sec \alpha = \frac{1}{\cos \alpha} = \sqrt{3}$.

(B) Since the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$,the direction ratios of the line $(3, -5, 2)$ must be perpendicular to the normal of the plane $(1, 3, -\alpha)$.
Therefore,$3(1) + (-5)(3) + 2(-\alpha) = 0$.
$3 - 15 - 2\alpha = 0 \Rightarrow -12 - 2\alpha = 0 \Rightarrow \alpha = -6$.
Also,the point $(2, 1, -2)$ on the line must satisfy the plane equation.
$2 + 3(1) - (-6)(-2) + \beta = 0$.
$2 + 3 - 12 + \beta = 0 \Rightarrow -7 + \beta = 0 \Rightarrow \beta = 7$.
Finally,the value of $\alpha^2 + \alpha\beta + \beta^2 = (-6)^2 + (-6)(7) + (7)^2 = 36 - 42 + 49 = 43$.

(C) The angle $\theta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by $\sin \theta = \left| \frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|} \right|$.
Here,the direction vector of the line is $\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$ and the normal vector of the plane is $\vec{n} = \hat{i} - 2\hat{j} - \lambda\hat{k}$.
Given $\theta = \cos^{-1}\left(\frac{2\sqrt{2}}{3}\right)$. Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = \sqrt{1 - \left(\frac{2\sqrt{2}}{3}\right)^2} = \sqrt{1 - \frac{8}{9}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.
Substituting the values into the formula:
$\frac{1}{3} = \left| \frac{(2)(1) + (1)(-2) + (-2)(-\lambda)}{\sqrt{2^2 + 1^2 + (-2)^2} \sqrt{1^2 + (-2)^2 + (-\lambda)^2}} \right|$
$\frac{1}{3} = \left| \frac{2 - 2 + 2\lambda}{\sqrt{4 + 1 + 4} \sqrt{1 + 4 + \lambda^2}} \right|$
$\frac{1}{3} = \left| \frac{2\lambda}{3 \sqrt{5 + \lambda^2}} \right|$
Squaring both sides:
$\frac{1}{9} = \frac{4\lambda^2}{9(5 + \lambda^2)}$
$5 + \lambda^2 = 4\lambda^2$
$3\lambda^2 = 5$
$\lambda^2 = \frac{5}{3} \Rightarrow \lambda = \sqrt{\frac{5}{3}}$.

(B) Let the line be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}=k$.
Any point $P$ on the line is given by $(2k+1, 3k-1, 4k+2)$.
Since $P$ lies on the plane $x+2y+3z=15$,we substitute the coordinates of $P$ into the plane equation:
$(2k+1) + 2(3k-1) + 3(4k+2) = 15$.
Expanding the terms: $2k + 1 + 6k - 2 + 12k + 6 = 15$.
Combining like terms: $20k + 5 = 15$,which gives $20k = 10$,so $k = \frac{1}{2}$.
Substituting $k = \frac{1}{2}$ back into the coordinates of $P$:
$P = (2(\frac{1}{2})+1, 3(\frac{1}{2})-1, 4(\frac{1}{2})+2) = (2, \frac{1}{2}, 4)$.
The distance of $P(2, \frac{1}{2}, 4)$ from the origin $(0, 0, 0)$ is $\sqrt{2^2 + (\frac{1}{2})^2 + 4^2} = \sqrt{4 + \frac{1}{4} + 16} = \sqrt{20 + \frac{1}{4}} = \sqrt{\frac{81}{4}} = \frac{9}{2}$ units.

(D) The normal vector $\bar{n_1}$ to plane $P_1$ is given by the cross product of its parallel vectors: $\bar{n_1} = (2 \hat{j}+3 \hat{k}) \times (4 \hat{j}-3 \hat{k}) = -18 \hat{i}$.
The normal vector $\bar{n_2}$ to plane $P_2$ is given by the cross product of its parallel vectors: $\bar{n_2} = (\hat{j}-\hat{k}) \times (3 \hat{i}+3 \hat{j}) = 3 \hat{i}-3 \hat{j}-3 \hat{k}$.
The vector $\bar{A}$ is parallel to the line of intersection,so $\bar{A}$ is perpendicular to both normals: $\bar{A} = \bar{n_1} \times \bar{n_2} = (-18 \hat{i}) \times (3 \hat{i}-3 \hat{j}-3 \hat{k}) = 54 \hat{j}-54 \hat{k}$.
We can simplify $\bar{A}$ to $\hat{j}-\hat{k}$ (or $-\hat{j}+\hat{k}$).
Let $\bar{v} = 2 \hat{i}+\hat{j}-2 \hat{k}$. The angle $\theta$ between $\bar{A}$ and $\bar{v}$ is given by $\cos \theta = \frac{|\bar{A} \cdot \bar{v}|}{|\bar{A}| |\bar{v}|}$.
$\bar{A} \cdot \bar{v} = (0)(2) + (1)(1) + (-1)(-2) = 1 + 2 = 3$.
$|\bar{A}| = \sqrt{0^2+1^2+(-1)^2} = \sqrt{2}$.
$|\bar{v}| = \sqrt{2^2+1^2+(-2)^2} = \sqrt{4+1+4} = 3$.
$\cos \theta = \frac{3}{\sqrt{2} \times 3} = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = \frac{\pi}{4}$. Since $\frac{\pi}{4}$ is not in the options,we consider the angle between $\bar{A} = -\hat{j}+\hat{k}$ and $\bar{v}$.
$\bar{A} \cdot \bar{v} = (0)(2) + (-1)(1) + (1)(-2) = -1 - 2 = -3$.
$\cos \theta = \frac{|-3|}{\sqrt{2} \times 3} = \frac{1}{\sqrt{2}}$.
Wait,if we take the angle between the vectors themselves (not the acute angle),$\cos \theta = \frac{-3}{3\sqrt{2}} = -\frac{1}{\sqrt{2}}$,which gives $\theta = \frac{3\pi}{4}$.

(A) Let the line be $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} = \lambda$.
Any point $P$ on the line is given by $(3\lambda + 2, 4\lambda - 1, 12\lambda + 2)$.
Since this point lies on the plane $x - y + z = 5$,we substitute the coordinates:
$(3\lambda + 2) - (4\lambda - 1) + (12\lambda + 2) = 5$.
$3\lambda + 2 - 4\lambda + 1 + 12\lambda + 2 = 5$.
$11\lambda + 5 = 5$.
$11\lambda = 0 \Rightarrow \lambda = 0$.
Thus,the point of intersection $P$ is $(2, -1, 2)$.
We need to find the distance between $P(2, -1, 2)$ and $Q(-1, -5, -10)$.
$PQ = \sqrt{(-1 - 2)^2 + (-5 - (-1))^2 + (-10 - 2)^2}$.
$PQ = \sqrt{(-3)^2 + (-4)^2 + (-12)^2}$.
$PQ = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$ units.

(A) The direction ratios of the line $\frac{x}{2} = \frac{y}{4} = \frac{z}{1}$ are $2, 4, 1$.
Since the required line is parallel to this line,its direction ratios are also $2, 4, 1$.
The equation of the line passing through $A(1, 3, 2)$ with direction ratios $2, 4, 1$ is $\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 2}{1} = \lambda$.
Any point $B$ on this line can be represented as $(2\lambda + 1, 4\lambda + 3, \lambda + 2)$.
Since point $B$ lies on the plane $3x + y + 2z = 5$,we substitute the coordinates of $B$ into the plane equation:
$3(2\lambda + 1) + (4\lambda + 3) + 2(\lambda + 2) = 5$.
$6\lambda + 3 + 4\lambda + 3 + 2\lambda + 4 = 5$.
$12\lambda + 10 = 5$.
$12\lambda = -5 \Rightarrow \lambda = -\frac{5}{12}$.
Substituting $\lambda = -\frac{5}{12}$ back into the coordinates of $B$:
$x = 2(-\frac{5}{12}) + 1 = -\frac{5}{6} + 1 = \frac{1}{6}$.
$y = 4(-\frac{5}{12}) + 3 = -\frac{5}{3} + 3 = \frac{4}{3}$.
$z = -\frac{5}{12} + 2 = \frac{19}{12}$.
Thus,the coordinates of point $B$ are $(\frac{1}{6}, \frac{4}{3}, \frac{19}{12})$.

(C) Let $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} = \lambda$.
Therefore, the coordinates of any point $P$ on the line are given by $P = (3\lambda + 2, 4\lambda - 1, 12\lambda + 2)$.
Since this point $P$ lies on the plane $x - y + z = 16$, we substitute the coordinates into the plane equation:
$(3\lambda + 2) - (4\lambda - 1) + (12\lambda + 2) = 16$.
$3\lambda + 2 - 4\lambda + 1 + 12\lambda + 2 = 16$.
$11\lambda + 5 = 16$.
$11\lambda = 11$, which gives $\lambda = 1$.
Substituting $\lambda = 1$ back into the coordinates of $P$, we get $P = (3(1) + 2, 4(1) - 1, 12(1) + 2) = (5, 3, 14)$.
Now, we find the distance between $P(5, 3, 14)$ and $Q(1, 6, 2)$ using the distance formula:
$PQ = \sqrt{(1-5)^2 + (6-3)^2 + (2-14)^2}$.
$PQ = \sqrt{(-4)^2 + (3)^2 + (-12)^2}$.
$PQ = \sqrt{16 + 9 + 144} = \sqrt{169} = 13 \text{ units}$.

(A) The line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $\bar{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is perpendicular to each of the normal vectors $\bar{n}_1 = 3 \hat{i}-\hat{j}+\hat{k}$ and $\bar{n}_2 = \hat{i}+4 \hat{j}-2 \hat{k}$.
Therefore,the line is parallel to the vector $\bar{n}_1 \times \bar{n}_2$.
Calculating the cross product:
$\bar{n}_1 \times \bar{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{vmatrix}$
$= \hat{i}((-1)(-2) - (1)(4)) - \hat{j}((3)(-2) - (1)(1)) + \hat{k}((3)(4) - (-1)(1))$
$= \hat{i}(2 - 4) - \hat{j}(-6 - 1) + \hat{k}(12 + 1)$
$= -2 \hat{i} + 7 \hat{j} + 13 \hat{k}$.

(A) Given $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$.
Using the identity $\cos ^{-1} a - \cos ^{-1} b = \cos ^{-1} (ab + \sqrt{1-a^2} \sqrt{1-b^2})$,we have:
$\cos ^{-1} \left( \frac{xy}{3} + \sqrt{1-x^2} \sqrt{1-\frac{y^2}{9}} \right) = \alpha$
$\frac{xy}{3} + \frac{\sqrt{1-x^2} \sqrt{9-y^2}}{3} = \cos \alpha$
$xy + \sqrt{1-x^2} \sqrt{9-y^2} = 3 \cos \alpha$
$xy - 3 \cos \alpha = -\sqrt{1-x^2} \sqrt{9-y^2}$
Squaring both sides:
$(xy - 3 \cos \alpha)^2 = (1-x^2)(9-y^2)$
$x^2y^2 - 6xy \cos \alpha + 9 \cos ^2 \alpha = 9 - y^2 - 9x^2 + x^2y^2$
$9x^2 - 6xy \cos \alpha + y^2 = 9 - 9 \cos ^2 \alpha$
$9x^2 - 6xy \cos \alpha + y^2 = 9(1 - \cos ^2 \alpha) = 9 \sin ^2 \alpha$.

(D) Let $f(\theta) = a \sec \theta - b \tan \theta$.
Then $f'(\theta) = a \sec \theta \tan \theta - b \sec^2 \theta = \sec \theta (a \tan \theta - b \sec \theta)$.
Setting $f'(\theta) = 0$,we get $a \tan \theta - b \sec \theta = 0$ (since $\sec \theta \neq 0$ for $0 < \theta < \frac{\pi}{2}$).
This implies $a \sin \theta = b$,or $\sin \theta = \frac{b}{a}$.
Since $\sin \theta = \frac{b}{a}$,we have $\cos \theta = \sqrt{1 - \frac{b^2}{a^2}} = \frac{\sqrt{a^2 - b^2}}{a}$.
Thus,$\sec \theta = \frac{a}{\sqrt{a^2 - b^2}}$ and $\tan \theta = \frac{b}{\sqrt{a^2 - b^2}}$.
The second derivative $f''(\theta) = a \sec \theta (1 + 2 \tan^2 \theta) > 0$,confirming a minimum at $\sin \theta = \frac{b}{a}$.
The minimum value is $f(\theta) = a \left( \frac{a}{\sqrt{a^2 - b^2}} \right) - b \left( \frac{b}{\sqrt{a^2 - b^2}} \right) = \frac{a^2 - b^2}{\sqrt{a^2 - b^2}} = \sqrt{a^2 - b^2}$.

(C) Let $f(x) = \sin x$.
Then $f^{\prime}(x) = \cos x$.
Here,$a = 60^{\circ}$ and $h = 10^{\prime \prime}$.
Since $1^{\circ} = 60^{\prime}$ and $1^{\prime} = 60^{\prime \prime}$,we have $1^{\circ} = 3600^{\prime \prime}$.
Thus,$h = \frac{10}{3600}^{\circ} = \frac{1}{360}^{\circ}$.
Converting $h$ to radians: $h = \frac{1}{360} \times 0.0175^{c} \approx 0.0000486^{c}$.
Now,$f(a) = \sin(60^{\circ}) = \frac{\sqrt{3}}{2} = \frac{1.732}{2} = 0.866$.
And $f^{\prime}(a) = \cos(60^{\circ}) = 0.5$.
Using the linear approximation formula $f(a+h) \approx f(a) + h f^{\prime}(a)$:
$\sin(60^{\circ} 0^{\prime} 10^{\prime \prime}) \approx 0.866 + (0.0000486 \times 0.5)$.
$\approx 0.866 + 0.0000243 = 0.8660243$.

(A) Given $\overline{OP} = \hat{a} \sin t + \hat{b} \cos t$.
$M = |\overline{OP}| = \sqrt{(\hat{a} \sin t + \hat{b} \cos t)^2}$.
Since $\hat{a}$ and $\hat{b}$ are unit vectors,$|\hat{a}| = 1$ and $|\hat{b}| = 1$.
$M^2 = \sin^2 t |\hat{a}|^2 + \cos^2 t |\hat{b}|^2 + 2 \sin t \cos t (\hat{a} \cdot \hat{b}) = \sin^2 t + \cos^2 t + \sin 2t (\hat{a} \cdot \hat{b}) = 1 + \sin 2t (\hat{a} \cdot \hat{b})$.
For $P$ to be farthest from $O$,$M$ must be maximum.
Since $\hat{a}$ and $\hat{b}$ form an acute angle,$\hat{a} \cdot \hat{b} > 0$.
Thus,$M$ is maximum when $\sin 2t = 1$,which implies $2t = \frac{\pi}{2}$,so $t = \frac{\pi}{4}$.
At $t = \frac{\pi}{4}$,$\sin t = \cos t = \frac{1}{\sqrt{2}}$.
$M = \sqrt{1 + 1(\hat{a} \cdot \hat{b})} = (1 + \hat{a} \cdot \hat{b})^{\frac{1}{2}}$.
$\hat{u} = \frac{\overline{OP}}{|\overline{OP}|} = \frac{\hat{a} \sin t + \hat{b} \cos t}{M} = \frac{\hat{a} \frac{1}{\sqrt{2}} + \hat{b} \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} |\hat{a} + \hat{b}|} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|}$.

(A) Let $\vec{a}$ and $\vec{b}$ be the adjacent sides of a parallelogram,where $\vec{a} = 2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\vec{b} = \hat{i}-2 \hat{j}-3 \hat{k}$.
One diagonal of the parallelogram is given by $\vec{c} = \vec{a} + \vec{b}$.
$\vec{c} = (2+1) \hat{i} + (-4-2) \hat{j} + (5-3) \hat{k} = 3 \hat{i} - 6 \hat{j} + 2 \hat{k}$.
The magnitude of $\vec{c}$ is $|\vec{c}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.
The unit vector parallel to the diagonal $\vec{c}$ is $\hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{3 \hat{i} - 6 \hat{j} + 2 \hat{k}}{7} = \frac{3}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{2}{7} \hat{k}$.

(B) Let $\vec{a} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{b} = \hat{i}+2\hat{j}+3\hat{k}$. The vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is given by $\vec{n} = \vec{a} \times \vec{b}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(3-1) + \hat{k}(2-1) = \hat{i}-2\hat{j}+\hat{k}$.
The projection of vector $\vec{c} = 2\hat{i}+\hat{j}+\hat{k}$ on $\vec{n}$ is given by $\left| \frac{\vec{c} \cdot \vec{n}}{|\vec{n}|} \right|$.
$\vec{c} \cdot \vec{n} = (2)(1) + (1)(-2) + (1)(1) = 2 - 2 + 1 = 1$.
$|\vec{n}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$.
Therefore,the magnitude of the projection is $\left| \frac{1}{\sqrt{6}} \right| = \frac{1}{\sqrt{6}}$.

(A) Given $\overline{a} = \hat{i} + \hat{j} + \hat{k}$ and $\overline{b} = \hat{j} - \hat{k}$. Let $\overline{r} = x \hat{i} + y \hat{j} + z \hat{k}$.
We are given $\overline{a} \times \overline{r} = \overline{b}$.
$\overline{a} \times \overline{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = (z-y) \hat{i} - (z-x) \hat{j} + (y-x) \hat{k}$.
Comparing this with $\overline{b} = 0 \hat{i} + 1 \hat{j} - 1 \hat{k}$,we get:
$z - y = 0 \implies z = y$ $(i)$
$x - z = 1 \implies x = z + 1$ (ii)
$y - x = -1$ (iii)
Also,$\overline{a} \cdot \overline{r} = 3 \implies x + y + z = 3$ (iv).
Substituting $y = z$ and $x = z + 1$ into (iv):
$(z + 1) + z + z = 3 \implies 3z + 1 = 3 \implies 3z = 2 \implies z = \frac{2}{3}$.
Thus,$y = \frac{2}{3}$ and $x = \frac{2}{3} + 1 = \frac{5}{3}$.
Therefore,$\overline{r} = \frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k}$.

(A) Given: $\overline{a} = 4 \hat{i} + 13 \hat{j} - 18 \hat{k}$,$\overline{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}$,and $\overline{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$.
We are given the relation $\overline{a} = m \overline{b} + n \overline{c}$.
Substituting the vectors,we get:
$4 \hat{i} + 13 \hat{j} - 18 \hat{k} = m(\hat{i} - 2 \hat{j} + 3 \hat{k}) + n(2 \hat{i} + 3 \hat{j} - 4 \hat{k})$
$4 \hat{i} + 13 \hat{j} - 18 \hat{k} = (m + 2n) \hat{i} + (-2m + 3n) \hat{j} + (3m - 4n) \hat{k}$
Comparing the coefficients of $\hat{i}, \hat{j},$ and $\hat{k}$ on both sides,we obtain the following system of linear equations:
$1) \ m + 2n = 4$
$2) \ -2m + 3n = 13$
Multiply equation $(1)$ by $2$ to get $2m + 4n = 8$. Adding this to equation $(2)$:
$(2m + 4n) + (-2m + 3n) = 8 + 13$
$7n = 21 \Rightarrow n = 3$
Substitute $n = 3$ into equation $(1)$:
$m + 2(3) = 4 \Rightarrow m + 6 = 4 \Rightarrow m = -2$
Thus,$m + n = -2 + 3 = 1$.

(A) Since the vectors $\bar{a}$,$\bar{b}$,and $\bar{c}$ are linearly dependent,their scalar triple product must be zero: $\bar{a} \cdot (\bar{b} \times \bar{c}) = 0$.
This is equivalent to the determinant being zero:
$\begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{vmatrix} = 0$.
Expanding the determinant:
$1(3\beta - 4\alpha) - 1(4\beta - 4) + 1(4\alpha - 3) = 0$.
$3\beta - 4\alpha - 4\beta + 4 + 4\alpha - 3 = 0$.
$-\beta + 1 = 0 \implies \beta = 1$.
Given $|\bar{c}| = \sqrt{3}$,we have $\sqrt{1^2 + \alpha^2 + \beta^2} = \sqrt{3}$.
$1 + \alpha^2 + \beta^2 = 3$.
Substituting $\beta = 1$:
$1 + \alpha^2 + 1 = 3 \implies \alpha^2 = 1 \implies \alpha = \pm 1$.
Checking the options,for $\alpha = 1$ and $\beta = 1$,the condition is satisfied.
Thus,the correct option is $A$.

(D) According to the given condition,the vector $(\bar{a}+\lambda \bar{b})$ is perpendicular to $\bar{c}$,so their dot product must be zero:
$(\bar{a}+\lambda \bar{b}) \cdot \bar{c} = 0$
First,calculate $\bar{a}+\lambda \bar{b}$:
$\bar{a}+\lambda \bar{b} = (2 \hat{i}+3 \hat{j}+2 \hat{k}) + \lambda(2 \hat{i}+\hat{j}-\hat{k}) = (2+2\lambda) \hat{i} + (3+\lambda) \hat{j} + (2-\lambda) \hat{k}$
Now,take the dot product with $\bar{c} = 3 \hat{i} - \hat{j} + 0 \hat{k}$:
$[(2+2\lambda) \hat{i} + (3+\lambda) \hat{j} + (2-\lambda) \hat{k}] \cdot (3 \hat{i} - \hat{j} + 0 \hat{k}) = 0$
$3(2+2\lambda) - 1(3+\lambda) + 0(2-\lambda) = 0$
$6 + 6\lambda - 3 - \lambda = 0$
$3 + 5\lambda = 0$
$5\lambda = -3$
$\lambda = \frac{-3}{5}$

(B) The area of a parallelogram with adjacent sides $\bar{a}$ and $\bar{b}$ is given by $|\bar{a} \times \bar{b}|$.
Given that $|\bar{a} \times \bar{b}| = 16$.
Now,the area of the parallelogram with adjacent sides $(3 \bar{a} + 2 \bar{b})$ and $(\bar{a} + 3 \bar{b})$ is given by the magnitude of their cross product:
$|(3 \bar{a} + 2 \bar{b}) \times (\bar{a} + 3 \bar{b})|$
$= |3(\bar{a} \times \bar{a}) + 9(\bar{a} \times \bar{b}) + 2(\bar{b} \times \bar{a}) + 6(\bar{b} \times \bar{b})|$
Since $\bar{a} \times \bar{a} = 0$ and $\bar{b} \times \bar{b} = 0$,and $\bar{b} \times \bar{a} = -(\bar{a} \times \bar{b})$:
$= |0 + 9(\bar{a} \times \bar{b}) - 2(\bar{a} \times \bar{b}) + 0|$
$= |7(\bar{a} \times \bar{b})|$
$= 7 |\bar{a} \times \bar{b}|$
$= 7 \times 16 = 112$ sq. units.

(C) Given: $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$.
According to the condition,the projection of $\overline{v}$ along $\overline{u}$ is equal to the projection of $\overline{w}$ along $\overline{u}$.
$\frac{\overline{v} \cdot \overline{u}}{|\overline{u}|} = \frac{\overline{w} \cdot \overline{u}}{|\overline{u}|}$
$\implies \overline{v} \cdot \overline{u} = \overline{w} \cdot \overline{u}$
$\implies (\overline{w} - \overline{v}) \cdot \overline{u} = 0$.
Now,consider $|\overline{u} - \overline{v} + \overline{w}|^2 = |\overline{u} + (\overline{w} - \overline{v})|^2$.
$= |\overline{u}|^2 + |\overline{w} - \overline{v}|^2 + 2\overline{u} \cdot (\overline{w} - \overline{v})$.
Since $(\overline{w} - \overline{v}) \cdot \overline{u} = 0$,the last term is $0$.
$= |\overline{u}|^2 + |\overline{w}|^2 + |\overline{v}|^2 - 2(\overline{w} \cdot \overline{v})$.
Since $\overline{v}$ and $\overline{w}$ are perpendicular,$\overline{w} \cdot \overline{v} = 0$.
$= (1)^2 + (3)^2 + (2)^2 - 0 = 1 + 9 + 4 = 14$.
Therefore,$|\overline{u} - \overline{v} + \overline{w}| = \sqrt{14}$.

(D) Given $|\overline{b}|=4, |\overline{c}|=1$ and $|\overline{b} \times \overline{c}|=\sqrt{15}$.
Let $\alpha$ be the angle between $\overline{b}$ and $\overline{c}$.
$|\overline{b} \times \overline{c}| = |\overline{b}||\overline{c}| \sin \alpha = \sqrt{15}$.
$4 \times 1 \times \sin \alpha = \sqrt{15} \implies \sin \alpha = \frac{\sqrt{15}}{4}$.
Then,$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{15}{16} = \frac{1}{16} \implies \cos \alpha = \frac{1}{4}$ (assuming $\alpha$ is acute).
Given $\overline{b} = 2\overline{c} + \lambda \overline{a}$,we have $\overline{b} - 2\overline{c} = \lambda \overline{a}$.
Squaring both sides:
$|\overline{b} - 2\overline{c}|^2 = |\lambda \overline{a}|^2$.
$|\overline{b}|^2 + 4|\overline{c}|^2 - 4(\overline{b} \cdot \overline{c}) = \lambda^2 |\overline{a}|^2$.
$|\overline{b}|^2 + 4|\overline{c}|^2 - 4(|\overline{b}||\overline{c}| \cos \alpha) = \lambda^2 |\overline{a}|^2$.
$16 + 4(1) - 4(4 \times 1 \times \frac{1}{4}) = \lambda^2 (2)^2$.
$16 + 4 - 4 = 4\lambda^2$.
$16 = 4\lambda^2 \implies \lambda^2 = 4 \implies \lambda = \pm 2$.
Since $4$ is an option,the value is $4$.

(B) Given that $\bar{a}+2 \bar{b}$ is collinear with $\bar{c}$,there exists a non-zero scalar $n$ such that $\bar{a}+2 \bar{b} = n \bar{c}$. (Equation $1$)
Similarly,since $\bar{b}+3 \bar{c}$ is collinear with $\bar{a}$,there exists a non-zero scalar $m$ such that $\bar{b}+3 \bar{c} = m \bar{a}$. (Equation $2$)
From Equation $2$,we have $\bar{b} = m \bar{a} - 3 \bar{c}$.
Substitute this into Equation $1$: $\bar{a} + 2(m \bar{a} - 3 \bar{c}) = n \bar{c}$.
This simplifies to: $\bar{a} + 2m \bar{a} - 6 \bar{c} = n \bar{c}$.
$(1 + 2m) \bar{a} = (n + 6) \bar{c}$.
Since $\bar{a}$ and $\bar{c}$ are non-zero and not collinear,the coefficients must be zero.
Therefore,$1 + 2m = 0 \Rightarrow m = -\frac{1}{2}$ and $n + 6 = 0 \Rightarrow n = -6$.
Substituting $n = -6$ into Equation $1$,we get $\bar{a} + 2 \bar{b} = -6 \bar{c}$.

(C) Let the vector be $\vec{v} = a \hat{i} + b \hat{j} + c \hat{k}$.
Since $\vec{v}$ is coplanar with $\vec{a} = 2 \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} + \hat{k}$,it must be of the form $\vec{v} = \lambda \vec{a} + \mu \vec{b} = (2\lambda + \mu) \hat{i} + (\lambda + \mu) \hat{j} + (\lambda + \mu) \hat{k}$.
Since $\vec{v}$ is orthogonal to $\vec{c} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$,we have $\vec{v} \cdot \vec{c} = 0$.
$3(2\lambda + \mu) + 2(\lambda + \mu) + 6(\lambda + \mu) = 0$.
$6\lambda + 3\mu + 2\lambda + 2\mu + 6\lambda + 6\mu = 14\lambda + 11\mu = 0$.
Let $\lambda = 11$,then $\mu = -14$.
$\vec{v} = (2(11) - 14) \hat{i} + (11 - 14) \hat{j} + (11 - 14) \hat{k} = 8 \hat{i} - 3 \hat{j} - 3 \hat{k}$.
The unit vector is $\pm \frac{8 \hat{i} - 3 \hat{j} - 3 \hat{k}}{\sqrt{64 + 9 + 9}} = \pm \frac{8 \hat{i} - 3 \hat{j} - 3 \hat{k}}{\sqrt{82}}$.
Comparing with the options,the correct choice is $\frac{-8 \hat{i} + 3 \hat{j} + 3 \hat{k}}{\sqrt{82}}$ which is option $C$.

(C) The magnitude of a vector remains invariant under the rotation of the coordinate system about the origin.
Given the components of vector $\bar{a}$ in the original system are $(1, 2p)$,its squared magnitude is:
$|\bar{a}|^2 = 1^2 + (2p)^2 = 1 + 4p^2$
In the new system,the components are $(1, p+1)$,so its squared magnitude is:
$|\bar{b}|^2 = 1^2 + (p+1)^2 = 1 + p^2 + 2p + 1 = p^2 + 2p + 2$
Since $|\bar{a}|^2 = |\bar{b}|^2$,we have:
$1 + 4p^2 = p^2 + 2p + 2$
$3p^2 - 2p - 1 = 0$
Factoring the quadratic equation:
$(3p + 1)(p - 1) = 0$
Thus,$p = -\frac{1}{3}$ or $p = 1$.

(D) The projection of vector $\overline{b}$ in the direction of vector $\overline{a}$ is given by the formula: $\text{Projection} = \frac{\overline{a} \cdot \overline{b}}{|\overline{a}|}$.
First,calculate the dot product $\overline{a} \cdot \overline{b} = (2)(1) + (3)(-1) + (-4)(-1) = 2 - 3 + 4 = 3$.
Next,calculate the magnitude of $\overline{a}$: $|\overline{a}| = \sqrt{2^2 + 3^2 + (-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$.
Therefore,the projection is $\frac{3}{\sqrt{29}}$.

(A) Since the vectors $p \hat{i}+\hat{j}+\hat{k}$,$\hat{i}+q \hat{j}+\hat{k}$,and $\hat{i}+\hat{j}+r \hat{k}$ are coplanar,their scalar triple product must be zero:
$\left|\begin{array}{lll}p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & r\end{array}\right|=0$
Expanding the determinant along the first row:
$p(qr-1) - 1(r-1) + 1(1-q) = 0$
$pqr - p - r + 1 + 1 - q = 0$
$pqr - p - q - r + 2 = 0$
Rearranging the terms to find the required value:
$pqr - (p+q+r) = -2$

(D) Let $\theta$ be the angle between $\bar{a}$ and $\bar{b}$.
Since $\overline{c}=\lambda(\overline{a} \times \overline{b})$,it implies $\overline{c}$ is perpendicular to both $\overline{a}$ and $\overline{b}$.
Thus,$\overline{c} \cdot \overline{a} = 0$ and $\overline{c} \cdot \overline{b} = 0$.
Given $|\overline{a}+\overline{b}+\overline{c}|=1$,squaring both sides gives $|\overline{a}+\overline{b}+\overline{c}|^2 = 1$.
Expanding this,we get $|\overline{a}|^2+|\overline{b}|^2+|\overline{c}|^2+2(\overline{a} \cdot \overline{b}+\overline{b} \cdot \overline{c}+\overline{c} \cdot \overline{a}) = 1$.
Substituting the values $|\overline{a}|^2 = \frac{1}{3}$,$|\overline{b}|^2 = \frac{1}{2}$,$|\overline{c}|^2 = \frac{1}{6}$ and the dot products involving $\overline{c}$ as $0$:
$\frac{1}{3} + \frac{1}{2} + \frac{1}{6} + 2(\overline{a} \cdot \overline{b}) + 0 + 0 = 1$.
$\frac{2+3+1}{6} + 2|\overline{a}||\overline{b}| \cos \theta = 1$.
$1 + 2(\frac{1}{\sqrt{3}})(\frac{1}{\sqrt{2}}) \cos \theta = 1$.
$2(\frac{1}{\sqrt{6}}) \cos \theta = 0$.
Therefore,$\cos \theta = 0$,which means $\theta = \frac{\pi}{2}$.

(C) Given that the projection of $\overline{v}$ along $\overline{u}$ is equal to the projection of $\overline{w}$ along $\overline{u}$.
Since $|\overline{u}|=1$,the projection formula $\frac{\overline{v} \cdot \overline{u}}{|\overline{u}|}$ simplifies to $\overline{v} \cdot \overline{u}$.
Thus,$\overline{v} \cdot \overline{u} = \overline{w} \cdot \overline{u} \Rightarrow (\overline{v} - \overline{w}) \cdot \overline{u} = 0 \dots (i)$.
Also,$\overline{v}$ and $\overline{w}$ are perpendicular,so $\overline{v} \cdot \overline{w} = 0 \dots (ii)$.
We need to find $|\overline{u} - \overline{v} + \overline{w}|^2$.
$|\overline{u} - \overline{v} + \overline{w}|^2 = |\overline{u}|^2 + |-\overline{v}|^2 + |\overline{w}|^2 - 2(\overline{u} \cdot \overline{v}) + 2(\overline{u} \cdot \overline{w}) - 2(\overline{v} \cdot \overline{w})$.
Substituting the values $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$ and using $(i)$ and $(ii)$:
$|\overline{u} - \overline{v} + \overline{w}|^2 = 1^2 + 2^2 + 3^2 - 2(\overline{u} \cdot \overline{v} - \overline{u} \cdot \overline{w}) - 2(0)$.
Since $\overline{u} \cdot \overline{v} = \overline{u} \cdot \overline{w}$,the term $2(\overline{u} \cdot \overline{v} - \overline{u} \cdot \overline{w}) = 0$.
Therefore,$|\overline{u} - \overline{v} + \overline{w}|^2 = 1 + 4 + 9 = 14$.
Hence,$|\overline{u} - \overline{v} + \overline{w}| = \sqrt{14}$.

(A) Given that $|\overline{a}|=2, |\overline{b}|=3, |\overline{c}|=5$ and the angle between each pair of vectors is $60^{\circ}$.
We know that $\overline{a} \cdot \overline{b} = |\overline{a}||\overline{b}| \cos 60^{\circ} = (2)(3)(\frac{1}{2}) = 3$.
Similarly,$\overline{b} \cdot \overline{c} = |\overline{b}||\overline{c}| \cos 60^{\circ} = (3)(5)(\frac{1}{2}) = \frac{15}{2}$.
And $\overline{c} \cdot \overline{a} = |\overline{c}||\overline{a}| \cos 60^{\circ} = (5)(2)(\frac{1}{2}) = 5$.
Now,$|\overline{a}+\overline{b}+\overline{c}|^2 = |\overline{a}|^2+|\overline{b}|^2+|\overline{c}|^2 + 2(\overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a})$.
Substituting the values: $|\overline{a}+\overline{b}+\overline{c}|^2 = 2^2 + 3^2 + 5^2 + 2(3 + \frac{15}{2} + 5)$.
$|\overline{a}+\overline{b}+\overline{c}|^2 = 4 + 9 + 25 + 2(\frac{6+15+10}{2}) = 38 + 31 = 69$.
Therefore,$|\overline{a}+\overline{b}+\overline{c}| = \sqrt{69}$.

(B) Let the given expression be $E = (\overline{a}-2 \overline{b}) \cdot \{(\overline{a} \times \overline{b}) \times (2 \overline{a}+\overline{b})\}$.
Using the vector triple product identity $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$,we have:
$(\overline{a} \times \overline{b}) \times (2 \overline{a}+\overline{b}) = -((2 \overline{a}+\overline{b}) \times (\overline{a} \times \overline{b}))$
$= -\{(2 \overline{a}+\overline{b}) \cdot \overline{b}) \overline{a} - ((2 \overline{a}+\overline{b}) \cdot \overline{a}) \overline{b}\}$
Since $\overline{a} \cdot \overline{a} = 1$ and $\overline{b} \cdot \overline{b} = 1$ (unit vectors) and $\overline{a} \cdot \overline{b} = \frac{1}{7\sqrt{10}}(3 \times 2 + 0 \times 3 + 1 \times (-6)) = 0$,the vectors are orthogonal.
Thus,$(2 \overline{a}+\overline{b}) \cdot \overline{b} = 2(\overline{a} \cdot \overline{b}) + \overline{b} \cdot \overline{b} = 0 + 1 = 1$.
And $(2 \overline{a}+\overline{b}) \cdot \overline{a} = 2(\overline{a} \cdot \overline{a}) + \overline{b} \cdot \overline{a} = 2(1) + 0 = 2$.
So,the expression becomes $- \{1 \cdot \overline{a} - 2 \cdot \overline{b}\} = 2 \overline{b} - \overline{a}$.
Now,$E = (\overline{a}-2 \overline{b}) \cdot (2 \overline{b} - \overline{a}) = -(\overline{a}-2 \overline{b}) \cdot (\overline{a}-2 \overline{b}) = -|\overline{a}-2 \overline{b}|^2$.
$|\overline{a}-2 \overline{b}|^2 = |\overline{a}|^2 + 4|\overline{b}|^2 - 4(\overline{a} \cdot \overline{b}) = 1 + 4(1) - 0 = 5$.
Therefore,$E = -5$.

(B) Let $\theta$ be the angle between $\overline{a}$ and $\overline{b}$.
Since $\overline{a}$ and $\overline{b}$ are unit vectors,$|\overline{a}| = 1$ and $|\overline{b}| = 1$.
Given that $\overline{c} = \overline{a} + 2\overline{b}$ and $\overline{d} = 5\overline{a} - 4\overline{b}$ are perpendicular,their dot product is zero:
$\overline{c} \cdot \overline{d} = 0$
$(\overline{a} + 2\overline{b}) \cdot (5\overline{a} - 4\overline{b}) = 0$
$5(\overline{a} \cdot \overline{a}) - 4(\overline{a} \cdot \overline{b}) + 10(\overline{b} \cdot \overline{a}) - 8(\overline{b} \cdot \overline{b}) = 0$
$5|\overline{a}|^2 + 6(\overline{a} \cdot \overline{b}) - 8|\overline{b}|^2 = 0$
Since $|\overline{a}| = 1$ and $|\overline{b}| = 1$,and $\overline{a} \cdot \overline{b} = |\overline{a}||\overline{b}| \cos \theta = \cos \theta$:
$5(1) + 6 \cos \theta - 8(1) = 0$
$6 \cos \theta - 3 = 0$
$6 \cos \theta = 3$
$\cos \theta = \frac{3}{6} = \frac{1}{2}$
$\theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$.

(B) Given: $|\bar{a}|=1, |\bar{b}|=4$ and $\bar{a} \cdot \bar{b}=2$.
We know that $|\bar{a} \times \bar{b}|^2 = |\bar{a}|^2 |\bar{b}|^2 - (\bar{a} \cdot \bar{b})^2$.
$|\bar{a} \times \bar{b}|^2 = (1)^2(4)^2 - (2)^2 = 16 - 4 = 12$.
Given $\bar{c} = 2(\bar{a} \times \bar{b}) - 3\bar{b}$.
Since $(\bar{a} \times \bar{b}) \perp \bar{b}$,the vectors $2(\bar{a} \times \bar{b})$ and $-3\bar{b}$ are orthogonal.
$|\bar{c}|^2 = |2(\bar{a} \times \bar{b})|^2 + |-3\bar{b}|^2 = 4|\bar{a} \times \bar{b}|^2 + 9|\bar{b}|^2$.
$|\bar{c}|^2 = 4(12) + 9(16) = 48 + 144 = 192$.
$|\bar{c}| = \sqrt{192} = 8\sqrt{3}$.
Now,$\bar{b} \cdot \bar{c} = \bar{b} \cdot (2(\bar{a} \times \bar{b}) - 3\bar{b}) = 2(\bar{b} \cdot (\bar{a} \times \bar{b})) - 3(\bar{b} \cdot \bar{b})$.
Since $\bar{b} \cdot (\bar{a} \times \bar{b}) = 0$,we have $\bar{b} \cdot \bar{c} = -3|\bar{b}|^2 = -3(16) = -48$.
Let $\theta$ be the angle between $\bar{b}$ and $\bar{c}$.
$\cos \theta = \frac{\bar{b} \cdot \bar{c}}{|\bar{b}| |\bar{c}|} = \frac{-48}{4 \times 8\sqrt{3}} = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$.
$\theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6}$.

(A) We are given the vector triple product identity: $(\overline{a} \times \overline{b}) \times \overline{c} = (\overline{a} \cdot \overline{c}) \overline{b} - (\overline{b} \cdot \overline{c}) \overline{a}$.
Comparing this with the given equation $(\overline{a} \times \overline{b}) \times \overline{c} = \frac{1}{3}|\overline{b}||\overline{c}| \overline{a}$,we note that the coefficient of $\overline{b}$ must be zero since there is no $\overline{b}$ term on the right side. Thus,$(\overline{a} \cdot \overline{c}) = 0$.
Equating the coefficients of $\overline{a}$,we get: $-(\overline{b} \cdot \overline{c}) = \frac{1}{3}|\overline{b}||\overline{c}|$.
Using the definition of the dot product,$\overline{b} \cdot \overline{c} = |\overline{b}||\overline{c}| \cos \theta$,we have:
$-|\overline{b}||\overline{c}| \cos \theta = \frac{1}{3}|\overline{b}||\overline{c}|$.
Since $\overline{b}$ and $\overline{c}$ are non-zero vectors,we can divide by $|\overline{b}||\overline{c}|$:
$-\cos \theta = \frac{1}{3} \implies \cos \theta = -\frac{1}{3}$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we find:
$\sin^2 \theta = 1 - (-\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}$.
Since $\theta$ is the angle between two vectors,$0 \le \theta \le \pi$,so $\sin \theta \ge 0$.
Therefore,$\sin \theta = \sqrt{\frac{8}{9}} = \frac{2 \sqrt{2}}{3}$.

(C) Let $\bar{r}$ be a vector in the plane of $\bar{a}$ and $\bar{b}$. Then,$\bar{r} = \bar{a} + m\bar{b}$ for some scalar $m$.
$\bar{r} = (\hat{i} + 2\hat{j} + \hat{k}) + m(\hat{i} - \hat{j} + \hat{k}) = (1+m)\hat{i} + (2-m)\hat{j} + (1+m)\hat{k}$.
The projection of $\bar{r}$ on $\bar{c}$ is given by $\frac{\bar{r} \cdot \bar{c}}{|\bar{c}|} = \frac{1}{\sqrt{3}}$.
First,calculate $|\bar{c}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$.
Now,$\bar{r} \cdot \bar{c} = (1+m)(1) + (2-m)(1) + (1+m)(-1) = 1+m + 2-m - 1-m = 2-m$.
Thus,$\frac{2-m}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow 2-m = 1 \Rightarrow m = 1$.
Substituting $m=1$ into the expression for $\bar{r}$:
$\bar{r} = (1+1)\hat{i} + (2-1)\hat{j} + (1+1)\hat{k} = 2\hat{i} + \hat{j} + 2\hat{k}$.
Checking the options,if we consider the projection as $\pm \frac{1}{\sqrt{3}}$,then $2-m = -1 \Rightarrow m = 3$.
For $m=3$,$\bar{r} = (1+3)\hat{i} + (2-3)\hat{j} + (1+3)\hat{k} = 4\hat{i} - \hat{j} + 4\hat{k}$.
This matches option $C$.

(B) Given that the angle between $\bar{a} \times \bar{b}$ and $\bar{c}$ is $\frac{\pi}{4}$.
Using the definition of the cross product magnitude,$|(\bar{a} \times \bar{b}) \times \bar{c}| = |\bar{a} \times \bar{b}| |\bar{c}| \sin(\frac{\pi}{4})$. ... $(i)$
First,calculate $\bar{a} \times \bar{b}$:
$\bar{a} \times \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i} - 2\hat{j} + \hat{k}$.
The magnitude is $|\bar{a} \times \bar{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Given $|\bar{c} - \bar{a}| = 2\sqrt{2}$. Squaring both sides:
$|\bar{c}|^2 + |\bar{a}|^2 - 2(\bar{a} \cdot \bar{c}) = (2\sqrt{2})^2 = 8$.
Since $|\bar{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = 3$ and $\bar{a} \cdot \bar{c} = |\bar{c}|$,we have:
$|\bar{c}|^2 + 3^2 - 2|\bar{c}| = 8$.
$|\bar{c}|^2 - 2|\bar{c}| + 9 = 8 \implies |\bar{c}|^2 - 2|\bar{c}| + 1 = 0$.
$(|\bar{c}| - 1)^2 = 0 \implies |\bar{c}| = 1$.
Substituting these values into equation $(i)$:
$|(\bar{a} \times \bar{b}) \times \bar{c}| = 3 \times 1 \times \sin(\frac{\pi}{4}) = 3 \times \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.

(C) First,we find the vectors $\overline{AB}$ and $\overline{CD}$:
$\overline{AB} = (1-2)\hat{i} + (-4-(-3))\hat{j} + (-2-0)\hat{k} = -\hat{i} - \hat{j} - 2\hat{k}$
$\overline{CD} = (7-4)\hat{i} + (0-6)\hat{j} + (10-8)\hat{k} = 3\hat{i} - 6\hat{j} + 2\hat{k}$
The vector projection of $\overline{AB}$ on $\overline{CD}$ is given by the formula:
$\text{Vector Projection} = (\overline{AB} \cdot \hat{CD}) \hat{CD} = (\overline{AB} \cdot \overline{CD}) \frac{\overline{CD}}{|\overline{CD}|^2}$
Calculate the dot product $\overline{AB} \cdot \overline{CD}$:
$\overline{AB} \cdot \overline{CD} = (-1)(3) + (-1)(-6) + (-2)(2) = -3 + 6 - 4 = -1$
Calculate the magnitude squared $|\overline{CD}|^2$:
$|\overline{CD}|^2 = 3^2 + (-6)^2 + 2^2 = 9 + 36 + 4 = 49$
Substitute these values into the formula:
$\text{Vector Projection} = (-1) \frac{3\hat{i} - 6\hat{j} + 2\hat{k}}{49} = -\frac{1}{49}(3\hat{i} - 6\hat{j} + 2\hat{k})$

(C) Let $\overline{d} = \overline{a} + \lambda \overline{b}$.
$\overline{d} = (2 \hat{i} + 3 \hat{j} + 2 \hat{k}) + \lambda(2 \hat{i} + \hat{j} - \hat{k})$
$\overline{d} = (2 + 2\lambda) \hat{i} + (3 + \lambda) \hat{j} + (2 - \lambda) \hat{k}$.
Since $\overline{d}$ is perpendicular to $\overline{c}$,their dot product must be zero,i.e.,$\overline{c} \cdot \overline{d} = 0$.
$(\hat{i} + 3 \hat{j}) \cdot [(2 + 2\lambda) \hat{i} + (3 + \lambda) \hat{j} + (2 - \lambda) \hat{k}] = 0$.
$1(2 + 2\lambda) + 3(3 + \lambda) + 0(2 - \lambda) = 0$.
$2 + 2\lambda + 9 + 3\lambda = 0$.
$5\lambda + 11 = 0$.
$\lambda = -\frac{11}{5}$.

(B) Let $\vec{a}$ be the vector joining points $A(-2,0,3)$ and $B(1,4,2)$.
$\vec{a} = (1 - (-2))\hat{i} + (4 - 0)\hat{j} + (2 - 3)\hat{k} = 3\hat{i} + 4\hat{j} - \hat{k}$.
Let $\vec{b}$ be the vector along the line with direction ratios $6, -2, 3$,so $\vec{b} = 6\hat{i} - 2\hat{j} + 3\hat{k}$.
The scalar projection of $\vec{a}$ on $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (3)(6) + (4)(-2) + (-1)(3) = 18 - 8 - 3 = 7$.
$|\vec{b}| = \sqrt{6^2 + (-2)^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.
Therefore,the scalar projection is $\frac{7}{7} = 1$.

(A) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,so $|\bar{a}| = 1$ and $|\bar{b}| = 1$.
Since $(\bar{a}+2 \bar{b})$ and $(5 \bar{a}-4 \bar{b})$ are perpendicular,their dot product is zero:
$(\bar{a}+2 \bar{b}) \cdot (5 \bar{a}-4 \bar{b}) = 0$
$5|\bar{a}|^2 - 4(\bar{a} \cdot \bar{b}) + 10(\bar{b} \cdot \bar{a}) - 8|\bar{b}|^2 = 0$
Substituting $|\bar{a}|^2 = 1$ and $|\bar{b}|^2 = 1$:
$5(1) + 6(\bar{a} \cdot \bar{b}) - 8(1) = 0$
$6(\bar{a} \cdot \bar{b}) - 3 = 0$
$\bar{a} \cdot \bar{b} = \frac{3}{6} = \frac{1}{2}$
Since $\bar{a} \cdot \bar{b} = |\bar{a}||\bar{b}| \cos \theta$,we have:
$1 \cdot 1 \cdot \cos \theta = \frac{1}{2}$
$\cos \theta = \frac{1}{2}$
$\theta = \frac{\pi}{3}$

(D) Let $\bar{r} = a\hat{i} + b\hat{j} + c\hat{k}$.
Since $\bar{r}$ is perpendicular to $\bar{q}$,we have $\bar{r} \cdot \bar{q} = 0$.
This implies $a - 2b + c = 0$ ... $(i)$.
Since $\bar{r}$ is coplanar with $\bar{p}$ and $\bar{q}$,the scalar triple product $[\bar{p} \ \bar{q} \ \bar{r}] = 0$.
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -2 & 1 \\ a & b & c \end{vmatrix} = 0$.
Expanding the determinant: $1(-2c - b) - 1(c - a) + 1(b + 2a) = 0$.
$-2c - b - c + a + b + 2a = 0 \Rightarrow 3a - 3c = 0 \Rightarrow a = c$ ... $(ii)$.
Substituting $(ii)$ into $(i)$: $a - 2b + a = 0 \Rightarrow 2a = 2b \Rightarrow a = b$.
Thus,$\bar{r} = a(\hat{i} + \hat{j} + \hat{k})$.
The unit vector in the direction of $\bar{r}$ is $\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.
The required vector has magnitude $5\sqrt{3}$,so $\text{Vector} = 5\sqrt{3} \times \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = 5(\hat{i} + \hat{j} + \hat{k})$.

(C) We know the vector triple product formula: $(\bar{a} \times \bar{b}) \times \bar{c} = (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{b} \cdot \bar{c}) \bar{a}$.
Given that $(\bar{a} \times \bar{b}) \times \bar{c} = -5 \bar{a} + 4 \bar{b}$.
Comparing the coefficients of $\bar{a}$ and $\bar{b}$,we get:
$-(\bar{b} \cdot \bar{c}) = -5 \implies \bar{b} \cdot \bar{c} = 5$
$(\bar{a} \cdot \bar{c}) = 4$
Now,we need to find $\bar{a} \times (\bar{b} \times \bar{c})$.
Using the vector triple product formula: $\bar{a} \times (\bar{b} \times \bar{c}) = (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c}$.
Substituting the known values $\bar{a} \cdot \bar{c} = 4$ and $\bar{a} \cdot \bar{b} = 3$:
$\bar{a} \times (\bar{b} \times \bar{c}) = 4 \bar{b} - 3 \bar{c}$.

(D) Given $\overline{r} \times \overline{a} = \overline{b} \times \overline{a} \implies (\overline{r} - \overline{b}) \times \overline{a} = \overline{0}$. This implies $(\overline{r} - \overline{b}) = k_1 \overline{a}$ for some scalar $k_1$.
Similarly,$\overline{r} \times \overline{b} = \overline{a} \times \overline{b} \implies (\overline{r} - \overline{a}) \times \overline{b} = \overline{0}$. This implies $(\overline{r} - \overline{a}) = k_2 \overline{b}$ for some scalar $k_2$.
From the first equation,$\overline{r} = \overline{b} + k_1 \overline{a} = (2\hat{j} - \hat{k}) + k_1(\hat{i} + \hat{j}) = k_1\hat{i} + (2+k_1)\hat{j} - \hat{k}$.
From the second equation,$\overline{r} = \overline{a} + k_2 \overline{b} = (\hat{i} + \hat{j}) + k_2(2\hat{j} - \hat{k}) = \hat{i} + (1+2k_2)\hat{j} - k_2\hat{k}$.
Comparing the components:
$k_1 = 1$
$2+k_1 = 1+2k_2 \implies 2+1 = 1+2k_2 \implies 2k_2 = 2 \implies k_2 = 1$
$-1 = -k_2 \implies k_2 = 1$.
Substituting $k_1=1$ into $\overline{r} = k_1\hat{i} + (2+k_1)\hat{j} - \hat{k}$,we get $\overline{r} = \hat{i} + 3\hat{j} - \hat{k}$.
The magnitude $|\overline{r}| = \sqrt{1^2 + 3^2 + (-1)^2} = \sqrt{1+9+1} = \sqrt{11}$.
Therefore,$\frac{\overline{r}}{|\overline{r}|} = \frac{\hat{i} + 3\hat{j} - \hat{k}}{\sqrt{11}}$.

(B) Given that $\overline{a} \cdot(\overline{b}+\overline{c})+\overline{b} \cdot(\overline{c}+\overline{a})+\overline{c} \cdot(\overline{a}+\overline{b})=0$.
Expanding the terms,we get $\overline{a} \cdot \overline{b} + \overline{a} \cdot \overline{c} + \overline{b} \cdot \overline{c} + \overline{b} \cdot \overline{a} + \overline{c} \cdot \overline{a} + \overline{c} \cdot \overline{b} = 0$.
Since the dot product is commutative,$\overline{a} \cdot \overline{b} = \overline{b} \cdot \overline{a}$,etc.
Thus,$2(\overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a}) = 0$,which implies $\overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a} = 0$.
We know that $|\overline{a}+\overline{b}+\overline{c}|^2 = |\overline{a}|^2 + |\overline{b}|^2 + |\overline{c}|^2 + 2(\overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a})$.
Substituting the given values $|\overline{a}|=1, |\overline{b}|=8, |\overline{c}|=4$ and the result from above:
$|\overline{a}+\overline{b}+\overline{c}|^2 = 1^2 + 8^2 + 4^2 + 2(0) = 1 + 64 + 16 = 81$.
Therefore,$|\overline{a}+\overline{b}+\overline{c}| = \sqrt{81} = 9$.

(A) Given $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=3 \hat{i}-2 \hat{j}+5 \hat{k}$.
First,calculate the sum and difference of the vectors:
$\bar{a}+\bar{b} = (\hat{i}+\hat{j}+\hat{k})+(3 \hat{i}-2 \hat{j}+5 \hat{k}) = 4 \hat{i}-\hat{j}+6 \hat{k}$
$\bar{a}-\bar{b} = (\hat{i}+\hat{j}+\hat{k})-(3 \hat{i}-2 \hat{j}+5 \hat{k}) = -2 \hat{i}+3 \hat{j}-4 \hat{k}$
The vector perpendicular to both $(\bar{a}+\bar{b})$ and $(\bar{a}-\bar{b})$ is given by their cross product:
$\vec{v} = (\bar{a}+\bar{b}) \times (\bar{a}-\bar{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 6 \\ -2 & 3 & -4 \end{vmatrix}$
$\vec{v} = \hat{i}((-1)(-4) - (6)(3)) - \hat{j}((4)(-4) - (6)(-2)) + \hat{k}((4)(3) - (-1)(-2))$
$\vec{v} = \hat{i}(4 - 18) - \hat{j}(-16 + 12) + \hat{k}(12 - 2) = -14 \hat{i} + 4 \hat{j} + 10 \hat{k}$
The magnitude of this vector is $|\vec{v}| = \sqrt{(-14)^2 + 4^2 + 10^2} = \sqrt{196 + 16 + 100} = \sqrt{312}$.
The required unit vector is $\frac{\vec{v}}{|\vec{v}|} = \frac{-14 \hat{i}+4 \hat{j}+10 \hat{k}}{\sqrt{312}}$.

(B) First,calculate the cross product $\overline{a} \times \overline{b}$:
$\overline{a} \times \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} = \hat{i}(28 - (-4)) - \hat{j}(7 - 6) + \hat{k}(-2 - 12) = 32 \hat{i} - \hat{j} - 14 \hat{k}$.
Since $\overline{d}$ is parallel to $\overline{a} \times \overline{b}$,we can write $\overline{d} = k(32 \hat{i} - \hat{j} - 14 \hat{k})$ for some scalar $k$.
Given $\overline{c} \cdot \overline{d} = 15$,where $\overline{c} = 2 \hat{i} - \hat{j} + 4 \hat{k}$:
$(2 \hat{i} - \hat{j} + 4 \hat{k}) \cdot k(32 \hat{i} - \hat{j} - 14 \hat{k}) = 15$
$k(64 + 1 - 56) = 15$
$k(9) = 15 \implies k = \frac{15}{9} = \frac{5}{3}$.
Thus,$\overline{d} = \frac{5}{3}(32 \hat{i} - \hat{j} - 14 \hat{k}) = \frac{160}{3} \hat{i} - \frac{5}{3} \hat{j} - \frac{70}{3} \hat{k}$.
Checking the options,we see that option $(B)$ is $90 \hat{i} - 3 \hat{j} - 42 \hat{k}$,which is $3(30 \hat{i} - \hat{j} - 14 \hat{k})$. Note that the vector $30 \hat{i} - \hat{j} - 14 \hat{k}$ is not parallel to $32 \hat{i} - \hat{j} - 14 \hat{k}$. However,testing option $(B)$ directly: $\overline{c} \cdot \overline{d} = (2)(90) + (-1)(-3) + (4)(-42) = 180 + 3 - 168 = 15$. Thus,option $(B)$ satisfies the condition.

(D) The magnitude of the cross product of two vectors $\bar{u}$ and $\bar{v}$ is given by the formula:
$|\bar{u} \times \bar{v}| = |\bar{u}| |\bar{v}| \sin(\theta)$
Given:
$|\bar{u}| = 8$
$|\bar{v}| = 12$
$\theta = 150^{\circ}$
Substituting the values into the formula:
$|\bar{u} \times \bar{v}| = 8 \times 12 \times \sin(150^{\circ})$
Since $\sin(150^{\circ}) = \sin(180^{\circ} - 30^{\circ}) = \sin(30^{\circ}) = \frac{1}{2}$,we have:
$|\bar{u} \times \bar{v}| = 8 \times 12 \times \frac{1}{2}$
$|\bar{u} \times \bar{v}| = 8 \times 6 = 48$
Therefore,the correct option is $D$.

(A) We are given the expression $E = |(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b})|^2+4(\bar{a} \cdot \bar{b})^2$.
Expanding the cross product term:
$(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b}) = (\bar{a} \times \bar{a}) + (\bar{a} \times \bar{b}) - (\bar{b} \times \bar{a}) - (\bar{b} \times \bar{b})$.
Since $\bar{a} \times \bar{a} = 0$ and $\bar{b} \times \bar{b} = 0$,and using the property $\bar{b} \times \bar{a} = -(\bar{a} \times \bar{b})$,we get:
$(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b}) = (\bar{a} \times \bar{b}) - (-(\bar{a} \times \bar{b})) = 2(\bar{a} \times \bar{b})$.
Now,substituting this into the expression:
$E = |2(\bar{a} \times \bar{b})|^2 + 4(\bar{a} \cdot \bar{b})^2 = 4|\bar{a} \times \bar{b}|^2 + 4(\bar{a} \cdot \bar{b})^2$.
Factoring out $4$:
$E = 4(|\bar{a} \times \bar{b}|^2 + (\bar{a} \cdot \bar{b})^2)$.
Using Lagrange's identity,$|\bar{a} \times \bar{b}|^2 + (\bar{a} \cdot \bar{b})^2 = |\bar{a}|^2 |\bar{b}|^2$.
Thus,$E = 4 |\bar{a}|^2 |\bar{b}|^2$.
Given $|\bar{a}|=4$ and $|\bar{b}|=3$:
$E = 4 \times (4)^2 \times (3)^2 = 4 \times 16 \times 9 = 576$.

(B) Let $\overline{r} = x \hat{i} + y \hat{j} + z \hat{k}$.
Given $\overline{r} \times \overline{a} = \overline{b}$,we have:
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 2 & 3 & 4 \end{vmatrix} = \hat{i} - 2 \hat{j} + \hat{k}$
$(4y - 3z) \hat{i} - (4x - 2z) \hat{j} + (3x - 2y) \hat{k} = \hat{i} - 2 \hat{j} + \hat{k}$
Comparing coefficients:
$4y - 3z = 1$ $(1)$
$4x - 2z = 2 \Rightarrow 2x - z = 1 \Rightarrow z = 2x - 1$ $(2)$
$3x - 2y = 1 \Rightarrow 2y = 3x - 1 \Rightarrow y = \frac{3x - 1}{2}$ $(3)$
Given $\overline{r} \cdot \overline{c} = 3$,we have $x + y - z = 3$ $(4)$
Substituting $(2)$ and $(3)$ into $(4)$:
$x + \frac{3x - 1}{2} - (2x - 1) = 3$
$2x + 3x - 1 - 4x + 2 = 6$
$x + 1 = 6 \Rightarrow x = 5$
Using $x=5$ in $(2)$ and $(3)$:
$z = 2(5) - 1 = 9$
$y = \frac{3(5) - 1}{2} = 7$
Thus,$\overline{r} = 5 \hat{i} + 7 \hat{j} + 9 \hat{k}$
$|\overline{r}| = \sqrt{5^2 + 7^2 + 9^2} = \sqrt{25 + 49 + 81} = \sqrt{155}$