MHT CET 2023 Mathematics Question Paper with Answer and Solution

(A) The sample space of the experiment is $S = \{1, 2, 3, 4, 5, 6\}$.
We determine the number of factors $(X)$ for each outcome:
$X(1) = 1$ (factor: $1$)
$X(2) = 2$ (factors: $1, 2$)
$X(3) = 2$ (factors: $1, 3$)
$X(4) = 3$ (factors: $1, 2, 4$)
$X(5) = 2$ (factors: $1, 5$)
$X(6) = 4$ (factors: $1, 2, 3, 6$)
Now,we calculate the probabilities for each value of $X$:
$P(X = 1) = P(\{1\}) = 1/6$
$P(X = 2) = P(\{2, 3, 5\}) = 3/6 = 1/2$
$P(X = 3) = P(\{4\}) = 1/6$
$P(X = 4) = P(\{6\}) = 1/6$
Thus,the probability distribution is as shown in option $A$.

(B) Let the position vectors of $A, B, C, D, E, F$ be $\overline{a}, \overline{b}, \overline{c}, \overline{d}, \overline{e}, \overline{f}$ respectively.
$\therefore \overline{d} = \frac{\overline{b} + \overline{c}}{2}, \overline{e} = \frac{\overline{c} + \overline{a}}{2}, \overline{f} = \frac{\overline{a} + \overline{b}}{2}$.
Now,$\overline{AD} + \frac{2}{3} \overline{BE} + \frac{1}{3} \overline{CF} = (\overline{d} - \overline{a}) + \frac{2}{3}(\overline{e} - \overline{b}) + \frac{1}{3}(\overline{f} - \overline{c})$.
Substituting the values:
$= \frac{\overline{b} + \overline{c}}{2} - \overline{a} + \frac{2}{3}\left(\frac{\overline{c} + \overline{a}}{2} - \overline{b}\right) + \frac{1}{3}\left(\frac{\overline{a} + \overline{b}}{2} - \overline{c}\right)$.
$= \frac{\overline{b} + \overline{c} - 2\overline{a}}{2} + \frac{\overline{c} + \overline{a} - 2\overline{b}}{3} + \frac{\overline{a} + \overline{b} - 2\overline{c}}{6}$.
$= \frac{3(\overline{b} + \overline{c} - 2\overline{a}) + 2(\overline{c} + \overline{a} - 2\overline{b}) + (\overline{a} + \overline{b} - 2\overline{c})}{6}$.
$= \frac{3\overline{b} + 3\overline{c} - 6\overline{a} + 2\overline{c} + 2\overline{a} - 4\overline{b} + \overline{a} + \overline{b} - 2\overline{c}}{6}$.
$= \frac{-3\overline{a} + 3\overline{c}}{6} = \frac{3(\overline{c} - \overline{a})}{6} = \frac{1}{2}(\overline{c} - \overline{a}) = \frac{1}{2} \overline{AC}$.

(C) Let $f(x) = x^3 + x - 1$.
Since $f(0) = -1 < 0$ and $f(1) = 1 > 0$,by the Intermediate Value Theorem,there exists at least one real root $c$ in the interval $(0, 1)$.
Alternatively,consider the derivative $f'(x) = 3x^2 + 1$.
Since $3x^2 + 1 > 0$ for all $x \in \mathbb{R}$,$f(x)$ is a strictly increasing function.
$A$ strictly increasing cubic function crosses the $X$-axis exactly once.
Therefore,the equation has exactly one real root.

(A) Given that $x, y, z$ are in $A.P.$
$\therefore 2y = x + z$ ... $(i)$
Also,$\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z$ are in $A.P.$
$\therefore 2 \tan ^{-1} y = \tan ^{-1} x + \tan ^{-1} z$
Using the formula $\tan ^{-1} a + \tan ^{-1} b = \tan ^{-1} \left( \frac{a+b}{1-ab} \right)$:
$\tan ^{-1} \left( \frac{2y}{1-y^2} \right) = \tan ^{-1} \left( \frac{x+z}{1-xz} \right)$
$\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$
Substituting $x+z = 2y$ from $(i)$:
$\frac{2y}{1-y^2} = \frac{2y}{1-xz}$
This implies either $2y = 0$ (which leads to $x=y=z=0$) or $1-y^2 = 1-xz$,which gives $y^2 = xz$.
Since $x, y, z$ are in both $A.P.$ and $G.P.$,we must have $x=y=z$.

(C) Let $P_0$ be the initial population and $P$ be the population at time $t$.
Given $\frac{dP}{dt} = kP$,where $k > 0$.
Separating variables and integrating,we get $\ln P = kt + C$.
At $t = 0$,$P = P_0$,so $C = \ln P_0$.
Thus,$\ln \left( \frac{P}{P_0} \right) = kt$.
Given that at $t = 6$,$P = 2P_0$,we have $\ln(2) = 6k$,so $k = \frac{\ln 2}{6}$.
At $t = 18$,$\ln \left( \frac{P}{P_0} \right) = \left( \frac{\ln 2}{6} \right) \times 18 = 3 \ln 2 = \ln(2^3) = \ln 8$.
Therefore,$\frac{P}{P_0} = 8$,which means the number of bacteria becomes $8$ times the initial number.

(D) Given that $x, y, z$ are in $G$.$P$.,we have $y^2 = xz$ $(i)$.
Also,$\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ are in $A$.$P$.,so $2 \tan^{-1} y = \tan^{-1} x + \tan^{-1} z$.
Applying the formula $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$,we get $\tan^{-1} \left( \frac{2y}{1-y^2} \right) = \tan^{-1} \left( \frac{x+z}{1-xz} \right)$.
This implies $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.
Since $y^2 = xz$,the denominators are equal,so $2y = x+z$ (ii).
From $(i)$ and (ii),$x, y, z$ are in both $A$.$P$. and $G$.$P$.,which implies $x = y = z$.

(C) Let $m$ be the mass of the substance at time $t$. The rate of decay is given by $\frac{dm}{dt} = -km$,where $k > 0$ is the decay constant.
Integrating the equation $\frac{dm}{m} = -k dt$,we get $\log m = -kt + c$.
At $t = 0$,$m = m_0$,so $c = \log m_0$.
Thus,$\log m = -kt + \log m_0$,which implies $\log(\frac{m}{m_0}) = -kt$.
Given the half-life $h$,at $t = h$,$m = \frac{m_0}{2}$.
Substituting these values: $\log(\frac{1}{2}) = -kh$,which gives $-\log 2 = -kh$,or $k = \frac{\log 2}{h}$.
The initial decay rate is the value of $\frac{dm}{dt}$ at $t = 0$.
$\frac{dm}{dt} = -km_0 = -(\frac{\log 2}{h})m_0 = -\frac{m_0}{h} \log 2$.

(C) Total number of cards $= 52$. Total number of queens $= 4$.
Probability of getting a queen in one draw,$p = \frac{4}{52} = \frac{1}{13}$.
Probability of not getting a queen,$q = 1 - p = \frac{12}{13}$.
Since the cards are drawn with replacement,this follows a binomial distribution $B(n, p)$ with $n = 2$ and $p = \frac{1}{13}$.
The mean of a binomial distribution is given by $E(X) = np$.
$E(X) = 2 \times \frac{1}{13} = \frac{2}{13}$.

(A) Let $AD$ be the median through vertex $A$ to side $BC$.
Since $D$ is the midpoint of $BC$,its coordinates are $D = \left(\frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2}\right) = \left(\frac{\lambda - 1}{2}, 4, \frac{\mu + 2}{2}\right)$.
The vector $\vec{AD}$ is given by $D - A = \left(\frac{\lambda - 1}{2} - 2, 4 - 3, \frac{\mu + 2}{2} - 5\right) = \left(\frac{\lambda - 5}{2}, 1, \frac{\mu - 8}{2}\right)$.
Since the median $AD$ is equally inclined to the coordinate axes,its direction ratios must be equal.
Therefore,$\frac{\lambda - 5}{2} = 1 = \frac{\mu - 8}{2}$.
Solving for $\lambda$: $\frac{\lambda - 5}{2} = 1 \implies \lambda - 5 = 2 \implies \lambda = 7$.
Solving for $\mu$: $\frac{\mu - 8}{2} = 1 \implies \mu - 8 = 2 \implies \mu = 10$.
Thus,$\lambda + \mu = 7 + 10 = 17$.

(C) The constraints are $x+y \leq 4$,$x \leq 2$,$y \leq 1$,$x+y \geq 1$,and $x, y \geq 0$.
To find the vertices,we solve the intersection points of the boundary lines:
$1$. Intersection of $x+y=1$ and $y=0$ gives $A(1,0)$.
$2$. Intersection of $x=2$ and $y=0$ gives $B(2,0)$.
$3$. Intersection of $x=2$ and $y=1$ gives $C(2,1)$.
$4$. Intersection of $x+y=1$ and $x=0$ gives $D(0,1)$.
Thus,the vertices of the feasible region are $(1,0), (2,0), (2,1), (0,1)$.

(B) The equation of the line passing through $A(3, 4, 1)$ and $B(5, 1, 6)$ is given by $\frac{x-3}{5-3} = \frac{y-4}{1-4} = \frac{z-1}{6-1}$.
This simplifies to $\frac{x-3}{2} = \frac{y-4}{-3} = \frac{z-1}{5} = k$.
Any point on the line is $(2k+3, -3k+4, 5k+1)$.
Since the line crosses the $XZ$-plane,the $y$-coordinate must be $0$.
So,$-3k+4 = 0$,which gives $k = \frac{4}{3}$.
Substituting $k = \frac{4}{3}$ into the coordinates:
$x = 2(\frac{4}{3}) + 3 = \frac{8}{3} + 3 = \frac{17}{3}$.
$z = 5(\frac{4}{3}) + 1 = \frac{20}{3} + 1 = \frac{23}{3}$.
Thus,the required point is $\left(\frac{17}{3}, 0, \frac{23}{3}\right)$.

(A) Let $A = (1, 2, 3)$ and $B = \left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$. The plane passes through the midpoint $M$ of $AB$ and is perpendicular to the line segment $AB$.
The midpoint $M$ is given by:
$M = \left( \frac{1 - \frac{7}{3}}{2}, \frac{2 - \frac{4}{3}}{2}, \frac{3 - \frac{1}{3}}{2} \right) = \left( \frac{-\frac{4}{3}}{2}, \frac{\frac{2}{3}}{2}, \frac{\frac{8}{3}}{2} \right) = \left( -\frac{2}{3}, \frac{1}{3}, \frac{4}{3} \right)$.
The direction ratios of the normal to the plane are the same as the direction ratios of the line $AB$:
$D.r.s = \left( 1 - (-\frac{7}{3}), 2 - (-\frac{4}{3}), 3 - (-\frac{1}{3}) \right) = \left( \frac{10}{3}, \frac{10}{3}, \frac{10}{3} \right)$.
We can take the normal vector as $\vec{n} = (1, 1, 1)$.
The equation of the plane is $1(x - x_0) + 1(y - y_0) + 1(z - z_0) = 0$ where $(x_0, y_0, z_0) = M = \left( -\frac{2}{3}, \frac{1}{3}, \frac{4}{3} \right)$:
$1(x + \frac{2}{3}) + 1(y - \frac{1}{3}) + 1(z - \frac{4}{3}) = 0$
$x + y + z + \frac{2}{3} - \frac{1}{3} - \frac{4}{3} = 0$
$x + y + z - 1 = 0 \Rightarrow x + y + z = 1$.
Checking the options:
$(A)$ $(1, -1, 1) \Rightarrow 1 - 1 + 1 = 1$. This point lies on the plane.

(C) Let $A = (1, 2, 0)$,$B = (1, 0, 2)$,and $C = (0, x, 1)$.
$\vec{AB} = (1-1)\hat{i} + (0-2)\hat{j} + (2-0)\hat{k} = -2\hat{j} + 2\hat{k}$.
$\vec{AC} = (0-1)\hat{i} + (x-2)\hat{j} + (1-0)\hat{k} = -\hat{i} + (x-2)\hat{j} + \hat{k}$.
The area of $\triangle ABC$ is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}| = \sqrt{6}$.
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -2 & 2 \\ -1 & x-2 & 1 \end{vmatrix} = \hat{i}(-2 - 2(x-2)) - \hat{j}(0 - (-2)) + \hat{k}(0 - 2) = (2-2x)\hat{i} - 2\hat{j} - 2\hat{k}$.
$|\vec{AB} \times \vec{AC}| = \sqrt{(2-2x)^2 + (-2)^2 + (-2)^2} = \sqrt{4(1-x)^2 + 4 + 4} = \sqrt{4(1-2x+x^2) + 8} = \sqrt{4x^2 - 8x + 12} = 2\sqrt{x^2 - 2x + 3}$.
Given $\frac{1}{2} |\vec{AB} \times \vec{AC}| = \sqrt{6}$,so $\sqrt{x^2 - 2x + 3} = \sqrt{6}$.
Squaring both sides: $x^2 - 2x + 3 = 6 \Rightarrow x^2 - 2x - 3 = 0$.
$(x-3)(x+1) = 0$,so $x = 3$ or $x = -1$.

(D) Let the points be $A = (5, -1, 4)$ and $B = (4, -1, 3)$.
The vector representing the line segment is $\vec{AB} = (4-5)\hat{i} + (-1-(-1))\hat{j} + (3-4)\hat{k} = -\hat{i} - \hat{k}$.
The magnitude of the vector is $|\vec{AB}| = \sqrt{(-1)^2 + 0^2 + (-1)^2} = \sqrt{2}$.
The normal vector to the plane $x+y+z=7$ is $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.
Let $\theta$ be the angle between the line segment $AB$ and the plane. The angle $\phi$ between the line segment and the normal to the plane is given by $\cos \phi = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{AB}| |\vec{n}|}$.
$\cos \phi = \frac{|(-1)(1) + (0)(1) + (-1)(1)|}{\sqrt{2} \sqrt{1^2+1^2+1^2}} = \frac{|-2|}{\sqrt{2} \sqrt{3}} = \frac{2}{\sqrt{6}}$.
Since $\theta$ is the angle with the plane,$\sin \theta = \cos \phi = \frac{2}{\sqrt{6}}$.
Then $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{4}{6}} = \sqrt{\frac{2}{6}} = \sqrt{\frac{1}{3}}$.
The length of the projection of the line segment on the plane is $|\vec{AB}| \cos \theta = \sqrt{2} \times \sqrt{\frac{1}{3}} = \sqrt{\frac{2}{3}}$.

(D) The normal vector to face $OPQ$ is given by the cross product of vectors $\vec{OP}$ and $\vec{OQ}$.
$\vec{n_1} = \vec{OP} \times \vec{OQ} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ -1 & 1 & 2 \end{vmatrix} = \hat{i}(2-3) - \hat{j}(4+3) + \hat{k}(2+1) = -\hat{i} - 7\hat{j} + 3\hat{k}$.
The normal vector to face $PQR$ is given by the cross product of vectors $\vec{PQ}$ and $\vec{PR}$.
$\vec{PQ} = (-1-2, 1-1, 2-3) = (-3, 0, -1)$ and $\vec{PR} = (1-2, 2-1, 1-3) = (-1, 1, -2)$.
$\vec{n_2} = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 0 & -1 \\ -1 & 1 & -2 \end{vmatrix} = \hat{i}(0+1) - \hat{j}(6-1) + \hat{k}(-3-0) = \hat{i} - 5\hat{j} - 3\hat{k}$.
The angle $\theta$ between the faces is the angle between their normal vectors:
$\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} = \frac{|(-1)(1) + (-7)(-5) + (3)(-3)|}{\sqrt{(-1)^2+(-7)^2+3^2} \sqrt{1^2+(-5)^2+(-3)^2}}$
$\cos \theta = \frac{|-1 + 35 - 9|}{\sqrt{1+49+9} \sqrt{1+25+9}} = \frac{25}{\sqrt{59} \sqrt{35}}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{25}{\sqrt{59} \sqrt{35}}\right)$.

(A) Let the vertices be $A(0,2,1)$,$B(-2,0,0)$,and $C(-2,0,2)$.
First,calculate the lengths of the sides of the triangle:
$a = BC = \sqrt{(-2 - (-2))^2 + (0 - 0)^2 + (2 - 0)^2} = \sqrt{0 + 0 + 4} = 2$.
$b = AC = \sqrt{(-2 - 0)^2 + (0 - 2)^2 + (2 - 1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
$c = AB = \sqrt{(-2 - 0)^2 + (0 - 2)^2 + (0 - 1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
The coordinates of the incentre $I$ are given by $\left(\frac{ax_A + bx_B + cx_C}{a+b+c}, \frac{ay_A + by_B + cy_C}{a+b+c}, \frac{az_A + bz_B + cz_C}{a+b+c}\right)$.
$I = \left(\frac{2(0) + 3(-2) + 3(-2)}{2+3+3}, \frac{2(2) + 3(0) + 3(0)}{2+3+3}, \frac{2(1) + 3(0) + 3(2)}{2+3+3}\right)$.
$I = \left(\frac{0 - 6 - 6}{8}, \frac{4 + 0 + 0}{8}, \frac{2 + 0 + 6}{8}\right)$.
$I = \left(-\frac{12}{8}, \frac{4}{8}, \frac{8}{8}\right) = \left(-\frac{3}{2}, \frac{1}{2}, 1\right)$.

(C) Since $\triangle ABC$ is right-angled at $A$,the vectors $\vec{AB}$ and $\vec{AC}$ are perpendicular,so their dot product is zero: $\vec{AB} \cdot \vec{AC} = 0$.
First,we find the vectors:
$\vec{AB} = (3-4)\hat{i} + (1-2)\hat{j} + (8-x)\hat{k} = -\hat{i} - \hat{j} + (8-x)\hat{k}$.
$\vec{AC} = (2-4)\hat{i} + (-1-2)\hat{j} + (2-x)\hat{k} = -2\hat{i} - 3\hat{j} + (2-x)\hat{k}$.
Now,compute the dot product:
$(-1)(-2) + (-1)(-3) + (8-x)(2-x) = 0$.
$2 + 3 + (16 - 8x - 2x + x^2) = 0$.
$5 + 16 - 10x + x^2 = 0$.
$x^2 - 10x + 21 = 0$.
Factoring the quadratic equation:
$(x-3)(x-7) = 0$.
Thus,$x = 3$ or $x = 7$.
Given the options,the correct value is $3$.

(A) Given relations are $l-5m+3n=0$ and $7l^2+5m^2-3n^2=0$.
From the first equation,$l = 5m - 3n$.
Substituting this into the second equation: $7(5m-3n)^2 + 5m^2 - 3n^2 = 0$.
$7(25m^2 - 30mn + 9n^2) + 5m^2 - 3n^2 = 0$.
$175m^2 - 210mn + 63n^2 + 5m^2 - 3n^2 = 0$.
$180m^2 - 210mn + 60n^2 = 0$.
Dividing by $30$,we get $6m^2 - 7mn + 2n^2 = 0$.
Factoring the quadratic: $(3m - 2n)(2m - n) = 0$.
Case $1$: $3m = 2n \Rightarrow m = \frac{2n}{3}$. Then $l = 5(\frac{2n}{3}) - 3n = \frac{10n-9n}{3} = \frac{n}{3}$.
Using $l^2 + m^2 + n^2 = 1$: $(\frac{n}{3})^2 + (\frac{2n}{3})^2 + n^2 = 1 \Rightarrow \frac{n^2}{9} + \frac{4n^2}{9} + n^2 = 1 \Rightarrow \frac{14n^2}{9} = 1 \Rightarrow n = \frac{3}{\sqrt{14}}$.
Then $l = \frac{1}{\sqrt{14}}$ and $m = \frac{2}{\sqrt{14}}$.
$l+m+n = \frac{1+2+3}{\sqrt{14}} = \frac{6}{\sqrt{14}}$.
Case $2$: $2m = n \Rightarrow m = \frac{n}{2}$. Then $l = 5(\frac{n}{2}) - 3n = \frac{5n-6n}{2} = -\frac{n}{2}$.
Using $l^2 + m^2 + n^2 = 1$: $(-\frac{n}{2})^2 + (\frac{n}{2})^2 + n^2 = 1 \Rightarrow \frac{n^2}{4} + \frac{n^2}{4} + n^2 = 1 \Rightarrow \frac{6n^2}{4} = 1 \Rightarrow n^2 = \frac{2}{3} \Rightarrow n = \sqrt{\frac{2}{3}}$.
Then $l = -\frac{1}{\sqrt{6}}$ and $m = \frac{1}{\sqrt{6}}$.
$l+m+n = \frac{-1+1+\sqrt{4}}{\sqrt{6}} = \frac{2}{\sqrt{6}}$.
Thus,the possible values are $\frac{2}{\sqrt{6}}$ or $\frac{6}{\sqrt{14}}$.

(C) Since the line makes equal angles with the coordinate axes,its direction cosines $(l, m, n)$ are equal. Let $l = m = n = a$.
Since $l^2 + m^2 + n^2 = 1$,we have $3a^2 = 1$,so $a = \frac{1}{\sqrt{3}}$ (taking positive value as direction cosines are positive).
Thus,the direction ratios of the line are proportional to $(1, 1, 1)$.
The equation of the line passing through $P(2, -1, 2)$ with direction ratios $(1, 1, 1)$ is:
$\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = k$
So,any point on the line is given by $(k+2, k-1, k+2)$.
Since this point $Q$ lies on the plane $2x+y+z=9$,we substitute the coordinates into the plane equation:
$2(k+2) + (k-1) + (k+2) = 9$
$2k + 4 + k - 1 + k + 2 = 9$
$4k + 5 = 9$
$4k = 4 \Rightarrow k = 1$
Substituting $k=1$ back into the coordinates of $Q$,we get $Q(1+2, 1-1, 1+2) = Q(3, 0, 3)$.
The length of the line segment $PQ$ is the distance between $P(2, -1, 2)$ and $Q(3, 0, 3)$:
$PQ = \sqrt{(3-2)^2 + (0 - (-1))^2 + (3-2)^2}$
$PQ = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

(B) Given equations are $l+m+n=0$ and $2l^2+2m^2-n^2=0$.
From the first equation,$n = -(l+m)$.
Substituting this into the second equation:
$2l^2 + 2m^2 - (-(l+m))^2 = 0$
$2l^2 + 2m^2 - (l^2 + m^2 + 2lm) = 0$
$l^2 + m^2 - 2lm = 0$
$(l-m)^2 = 0 \Rightarrow l=m$.
If $l=m$,then $n = -(l+l) = -2l$.
The direction ratios of the lines are proportional to $(l, l, -2l)$,which simplifies to $(1, 1, -2)$.
Since the direction ratios are the same for both lines,the lines are parallel.
The angle $\theta$ between parallel lines is $0^{\circ}$ or $180^{\circ}$.
Given the options,$180^{\circ}$ is the correct choice.

(A) The given Cartesian equation of the line is $6x-2=3y+1=2z-2$.
We rewrite this in the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ by factoring out the coefficients of $x, y, z$:
$6(x-\frac{1}{3})=3(y+\frac{1}{3})=2(z-1)$.
Dividing the entire equation by the least common multiple of $6, 3, 2$,which is $6$,we get:
$\frac{x-\frac{1}{3}}{1}=\frac{y+\frac{1}{3}}{2}=\frac{z-1}{3}$.
This line passes through the point $A(\frac{1}{3}, -\frac{1}{3}, 1)$ and has direction ratios proportional to $\vec{v} = \hat{i}+2\hat{j}+3\hat{k}$.
The vector equation of a line passing through point $\vec{a}$ with direction $\vec{v}$ is $\vec{r} = \vec{a} + \lambda \vec{v}$.
Substituting the values,we get $\vec{r} = (\frac{1}{3}\hat{i} - \frac{1}{3}\hat{j} + \hat{k}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$.

(D) Let the point be $P$ with position vector $\vec{\alpha} = \hat{i} - 2\hat{j} - 6\hat{k}$.
Let the line pass through point $A$ with position vector $\vec{a} = 2\hat{i} - 3\hat{j} - 4\hat{k}$ and be parallel to vector $\vec{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.
The distance $d$ of point $P$ from the line is given by $d = \frac{|(\vec{\alpha} - \vec{a}) \times \vec{b}|}{|\vec{b}|}$.
First,calculate $\vec{\alpha} - \vec{a} = (1-2)\hat{i} + (-2 - (-3))\hat{j} + (-6 - (-4))\hat{k} = -\hat{i} + \hat{j} - 2\hat{k}$.
Next,calculate the cross product $(\vec{\alpha} - \vec{a}) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 6 & 3 & -4 \end{vmatrix} = \hat{i}(-4 - (-6)) - \hat{j}(4 - (-12)) + \hat{k}(-3 - 6) = 2\hat{i} - 16\hat{j} - 9\hat{k}$.
The magnitude is $|(\vec{\alpha} - \vec{a}) \times \vec{b}| = \sqrt{2^2 + (-16)^2 + (-9)^2} = \sqrt{4 + 256 + 81} = \sqrt{341}$.
The magnitude of $\vec{b}$ is $|\vec{b}| = \sqrt{6^2 + 3^2 + (-4)^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$.
Therefore,the distance $d = \frac{\sqrt{341}}{\sqrt{61}} = \sqrt{\frac{341}{61}}$ units.

(B) The given equation of the line is $2x+4=3y+1=6z-3$.
To convert this into the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$,we rewrite the equation as:
$2(x+2)=3(y+\frac{1}{3})=6(z-\frac{1}{2})$.
Dividing throughout by the least common multiple of $2, 3, 6$,which is $6$,we get:
$\frac{2(x+2)}{6}=\frac{3(y+\frac{1}{3})}{6}=\frac{6(z-\frac{1}{2})}{6} \Rightarrow \frac{x+2}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{1}{2}}{1}$.
Thus,the line passes through the point $(-2, -\frac{1}{3}, \frac{1}{2})$ and has direction ratios $(3, 2, 1)$.
The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$.
Therefore,the vector equation is $\overline{r}=\left(-2 \hat{i}-\frac{1}{3} \hat{j}+\frac{1}{2} \hat{k}\right)+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})$.

(D) The given lines are $L_1: \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $L_2: \frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$.
Two lines are coplanar and intersect if the shortest distance between them is $0$.
The condition for intersection is given by the determinant of the vector connecting points on the lines and the direction vectors of the lines being $0$.
Let $(x_1, y_1, z_1) = (1, -1, 1)$ and $(x_2, y_2, z_2) = (-2, 1, -1)$.
The direction vectors are $(a_1, b_1, c_1) = (3, 2, 5)$ and $(a_2, b_2, c_2) = (4, 3, 2)$.
We calculate the determinant:
$\Delta = \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = \begin{vmatrix} -3 & 2 & -2 \\ 3 & 2 & 5 \\ 4 & 3 & 2 \end{vmatrix}$
$\Delta = -3(2 \times 2 - 5 \times 3) - 2(3 \times 2 - 5 \times 4) - 2(3 \times 3 - 2 \times 4)$
$\Delta = -3(4 - 15) - 2(6 - 20) - 2(9 - 8)$
$\Delta = -3(-11) - 2(-14) - 2(1)$
$\Delta = 33 + 28 - 2 = 59$.
Since $\Delta \neq 0$,the lines are skew and do not intersect.

(C) The required line passes through the point $A(1, 2, 3)$,so its position vector is $\overline{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.
Since the line is perpendicular to the lines with direction vectors $\overline{b_1} = 3\hat{i} + 2\hat{j} - 2\hat{k}$ and $\overline{b_2} = 2\hat{i} - 3\hat{j} + \hat{k}$,the direction vector $\overline{b}$ of the required line is given by the cross product $\overline{b_1} \times \overline{b_2}$.
$\overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -2 \\ 2 & -3 & 1 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(3 - (-4)) + \hat{k}(-9 - 4) = -4\hat{i} - 7\hat{j} - 13\hat{k}$.
The equation of the line is $\overline{r} = \overline{a} + \lambda\overline{b}$,which is $\overline{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-4\hat{i} - 7\hat{j} - 13\hat{k})$.

(D) The given lines are $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $L_2: \frac{x-3}{1}=\frac{y-k}{2}=\frac{z-0}{1}$.
For two lines to intersect,the shortest distance between them must be $0$.
This implies the determinant of the matrix formed by the difference of points and direction vectors must be $0$.
$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$
Substituting the values $(x_1, y_1, z_1) = (1, -1, 1)$,$(x_2, y_2, z_2) = (3, k, 0)$,$(a_1, b_1, c_1) = (2, 3, 4)$,and $(a_2, b_2, c_2) = (1, 2, 1)$:
$\begin{vmatrix} 3-1 & k-(-1) & 0-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0$
$\begin{vmatrix} 2 & k+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2(3(1) - 4(2)) - (k+1)(2(1) - 4(1)) - 1(2(2) - 3(1)) = 0$
$2(3-8) - (k+1)(2-4) - 1(4-3) = 0$
$2(-5) - (k+1)(-2) - 1(1) = 0$
$-10 + 2k + 2 - 1 = 0$
$2k - 9 = 0$
$k = \frac{9}{2}$

(A) First,rewrite the equations of the lines in the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$.
For the first line: $\frac{x-1}{-3}=\frac{y-2}{2p/7}=\frac{z-3}{2}$. The direction ratios are $\vec{v_1} = (-3, \frac{2p}{7}, 2)$.
For the second line: $\frac{x-1}{-3p/7}=\frac{y-5}{1}=\frac{z-6}{-5}$. The direction ratios are $\vec{v_2} = (-\frac{3p}{7}, 1, -5)$.
Since the lines are at right angles,their dot product must be zero: $\vec{v_1} \cdot \vec{v_2} = 0$.
$(-3)(-\frac{3p}{7}) + (\frac{2p}{7})(1) + (2)(-5) = 0$.
$\frac{9p}{7} + \frac{2p}{7} - 10 = 0$.
$\frac{11p}{7} = 10$.
$p = \frac{70}{11}$.

(B) Let the given point be $P(-2, 4, -5)$ and the line be $L: \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} = \lambda$.
Any point $Q$ on the line is given by $Q(3\lambda-3, 5\lambda+4, 6\lambda-8)$.
The vector $\vec{PQ} = (3\lambda-3 - (-2), 5\lambda+4-4, 6\lambda-8 - (-5)) = (3\lambda-1, 5\lambda, 6\lambda-3)$.
Since $PQ$ is perpendicular to the line with direction ratios $(3, 5, 6)$,their dot product is zero:
$3(3\lambda-1) + 5(5\lambda) + 6(6\lambda-3) = 0$
$9\lambda - 3 + 25\lambda + 36\lambda - 18 = 0$
$70\lambda - 21 = 0 \implies \lambda = \frac{21}{70} = \frac{3}{10}$.
Substituting $\lambda = \frac{3}{10}$ in $\vec{PQ}$:
$\vec{PQ} = (3(\frac{3}{10})-1, 5(\frac{3}{10}), 6(\frac{3}{10})-3) = (-\frac{1}{10}, \frac{15}{10}, -\frac{12}{10})$.
The distance $PQ = \sqrt{(-\frac{1}{10})^2 + (\frac{15}{10})^2 + (-\frac{12}{10})^2} = \sqrt{\frac{1+225+144}{100}} = \sqrt{\frac{370}{100}} = \sqrt{\frac{37}{10}}$ units.

(A) The given lines are $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $L_2: \vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda(\hat{i}+2\hat{j})$.
The point on $L_1$ is $A(-1, -2, -1)$ and its direction vector is $\vec{b_1} = 3\hat{i} + \hat{j} + 2\hat{k}$.
The point on $L_2$ is $B(2, -2, 3)$ and its direction vector is $\vec{b_2} = \hat{i} + 2\hat{j} + 0\hat{k}$.
The vector $\vec{AB} = (2 - (-1))\hat{i} + (-2 - (-2))\hat{j} + (3 - (-1))\hat{k} = 3\hat{i} + 0\hat{j} + 4\hat{k}$.
The cross product $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = \hat{i}(0-4) - \hat{j}(0-2) + \hat{k}(6-1) = -4\hat{i} + 2\hat{j} + 5\hat{k}$.
The magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-4)^2 + 2^2 + 5^2} = \sqrt{16 + 4 + 25} = \sqrt{45} = 3\sqrt{5}$.
The shortest distance $d = \frac{|\vec{AB} \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} = \frac{|(3\hat{i} + 0\hat{j} + 4\hat{k}) \cdot (-4\hat{i} + 2\hat{j} + 5\hat{k})|}{3\sqrt{5}} = \frac{|-12 + 0 + 20|}{3\sqrt{5}} = \frac{8}{3\sqrt{5}}$.

(A) The line passes through the point $(1, 2, 3)$ and is parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$.
Let the normal vectors to the planes be $\vec{n_1} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{n_2} = 3\hat{i} + \hat{j} + \hat{k}$.
The direction vector $\vec{v}$ of the line is parallel to the cross product of the normals: $\vec{v} = \vec{n_1} \times \vec{n_2}$.
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix} = \hat{i}(-1 - 2) - \hat{j}(1 - 6) + \hat{k}(1 + 3) = -3\hat{i} + 5\hat{j} + 4\hat{k}$.
The equation of the line passing through $(x_1, y_1, z_1)$ with direction vector $(a, b, c)$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Substituting the values,we get $\frac{x-1}{-3} = \frac{y-2}{5} = \frac{z-3}{4}$.

(B) Let the direction ratios of the required line be $a, b, c$.
Since the line is perpendicular to the lines with direction ratios $(1, 2, 3)$ and $(-3, 2, 5)$,we have:
$a + 2b + 3c = 0$ --- $(i)$
$-3a + 2b + 5c = 0$ --- $(ii)$
Using the cross-multiplication method to solve for $a, b, c$:
$\frac{a}{(2)(5) - (3)(2)} = \frac{b}{(3)(-3) - (1)(5)} = \frac{c}{(1)(2) - (2)(-3)}$
$\frac{a}{10 - 6} = \frac{b}{-9 - 5} = \frac{c}{2 + 6}$
$\frac{a}{4} = \frac{b}{-14} = \frac{c}{8}$
Dividing by $2$,we get the direction ratios as $(2, -7, 4)$.
The line passes through the point $(-1, 3, -2)$.
Therefore,the equation of the line is $\frac{x - (-1)}{2} = \frac{y - 3}{-7} = \frac{z - (-2)}{4}$,which simplifies to $\frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}$.

(B) The equation of line $L_1$ is given by $\vec{r} = 3 \hat{i} + \lambda(-\hat{i} + \hat{j} + \hat{k})$.
The equation of line $L_2$ is given by $\vec{r} = (\hat{i} + \hat{j}) + \mu(\hat{i} + \hat{k})$.
For the point of intersection,the position vectors must be equal:
$3 \hat{i} - \lambda \hat{i} + \lambda \hat{j} + \lambda \hat{k} = \hat{i} + \hat{j} + \mu \hat{i} + \mu \hat{k}$
Comparing the coefficients of $\hat{i}, \hat{j}, \hat{k}$:
For $\hat{j}$: $\lambda = 1$.
For $\hat{k}$: $\lambda = \mu$,so $\mu = 1$.
For $\hat{i}$: $3 - \lambda = 1 + \mu \Rightarrow 3 - 1 = 1 + 1 \Rightarrow 2 = 2$,which is consistent.
Substituting $\lambda = 1$ into the equation of $L_1$:
$\vec{r} = 3 \hat{i} + 1(-\hat{i} + \hat{j} + \hat{k}) = 2 \hat{i} + \hat{j} + \hat{k}$.
Thus,the position vector of the point of intersection is $2 \hat{i} + \hat{j} + \hat{k}$.

(C) The direction ratios (d.r.s.) of the line joining the points $P(2,1,-3)$ and $Q(-3,1,7)$ are given by $(x_2-x_1, y_2-y_1, z_2-z_1) = (-3-2, 1-1, 7-(-3)) = (-5, 0, 10)$.
The direction ratios of the line parallel to $\frac{x-1}{3}=\frac{y}{4}=\frac{z+3}{5}$ are $(3, 4, 5)$.
Let $\theta$ be the acute angle between these two lines. The formula for the angle between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is $\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$.
Substituting the values:
$\cos \theta = \left| \frac{(-5)(3) + (0)(4) + (10)(5)}{\sqrt{(-5)^2 + 0^2 + 10^2} \sqrt{3^2 + 4^2 + 5^2}} \right|$
$\cos \theta = \left| \frac{-15 + 0 + 50}{\sqrt{25 + 100} \sqrt{9 + 16 + 25}} \right|$
$\cos \theta = \frac{35}{\sqrt{125} \sqrt{50}} = \frac{35}{5\sqrt{5} \cdot 5\sqrt{2}} = \frac{35}{25\sqrt{10}} = \frac{7}{5\sqrt{10}}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{7}{5 \sqrt{10}}\right)$.

(A) Let the coordinates of any point $P$ on the line be $(3\lambda + 6, 2\lambda + 7, -2\lambda + 7)$.
Given point is $A(1, 2, 3)$.
The direction ratios of the line $AP$ are $(3\lambda + 6 - 1, 2\lambda + 7 - 2, -2\lambda + 7 - 3) = (3\lambda + 5, 2\lambda + 5, -2\lambda + 4)$.
Since $AP$ is perpendicular to the given line with direction ratios $(3, 2, -2)$,the dot product of their direction ratios must be zero:
$3(3\lambda + 5) + 2(2\lambda + 5) - 2(-2\lambda + 4) = 0$
$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0$
$17\lambda + 17 = 0$
$\lambda = -1$
Substituting $\lambda = -1$ into the coordinates of $P$:
$P = (3(-1) + 6, 2(-1) + 7, -2(-1) + 7) = (3, 5, 9)$.

(B) The lines $L_1$ and $L_2$ are parallel to the vectors $\vec{b}_1 = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b}_2 = \hat{i} + 2\hat{j} + 3\hat{k}$ respectively.
The unit vector perpendicular to both $L_1$ and $L_2$ is given by $\hat{n} = \pm \frac{\vec{b}_1 \times \vec{b}_2}{|\vec{b}_1 \times \vec{b}_2|}$.
First,calculate the cross product $\vec{b}_1 \times \vec{b}_2$:
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-4) - \hat{j}(9-2) + \hat{k}(6-1) = -\hat{i} - 7\hat{j} + 5\hat{k}$.
Next,calculate the magnitude $|\vec{b}_1 \times \vec{b}_2|$:
$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$.
Thus,the unit vector is $\hat{n} = \frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$.

(A) The shortest distance $d$ between two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ is given by $d = \frac{|(\vec{r_2}-\vec{r_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{ |\vec{b_1} \times \vec{b_2}| }$.
Here,$(x_1, y_1, z_1) = (1, 2, 3)$,$(x_2, y_2, z_2) = (2, 4, 5)$,$\vec{b_1} = 2\hat{i} + 3\hat{j} + \lambda\hat{k}$,and $\vec{b_2} = 1\hat{i} + 4\hat{j} + 5\hat{k}$.
$\vec{r_2}-\vec{r_1} = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$.
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & \lambda \\ 1 & 4 & 5 \end{vmatrix} = \hat{i}(15-4\lambda) - \hat{j}(10-\lambda) + \hat{k}(8-3) = (15-4\lambda)\hat{i} + (\lambda-10)\hat{j} + 5\hat{k}$.
The scalar triple product is $(\vec{r_2}-\vec{r_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 1(15-4\lambda) + 2(\lambda-10) + 2(5) = 15-4\lambda+2\lambda-20+10 = 5-2\lambda$.
The magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{(15-4\lambda)^2 + (\lambda-10)^2 + 5^2} = \sqrt{225-120\lambda+16\lambda^2 + \lambda^2-20\lambda+100 + 25} = \sqrt{17\lambda^2-140\lambda+350}$.
Given $d = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{|5-2\lambda|}{\sqrt{17\lambda^2-140\lambda+350}}$.
Squaring both sides: $\frac{1}{3} = \frac{(5-2\lambda)^2}{17\lambda^2-140\lambda+350} \Rightarrow 17\lambda^2-140\lambda+350 = 3(25-20\lambda+4\lambda^2) = 75-60\lambda+12\lambda^2$.
Rearranging: $5\lambda^2-80\lambda+275 = 0 \Rightarrow \lambda^2-16\lambda+55 = 0$.
Factoring: $(\lambda-5)(\lambda-11) = 0$,so $\lambda = 5$ or $\lambda = 11$.
The sum of possible values of $\lambda$ is $5+11 = 16$.

(D) Let $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} = \lambda$.
Any point $P$ on the line is given by $(3\lambda+6, 2\lambda+7, -2\lambda+7)$.
Let $A = (1, 2, 3)$.
The direction ratios of the line $AP$ are $(3\lambda+6-1, 2\lambda+7-2, -2\lambda+7-3)$, which simplifies to $(3\lambda+5, 2\lambda+5, -2\lambda+4)$.
Since $AP$ is perpendicular to the given line with direction ratios $(3, 2, -2)$, the dot product of their direction ratios must be zero:
$3(3\lambda+5) + 2(2\lambda+5) - 2(-2\lambda+4) = 0$.
$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0$.
$17\lambda + 17 = 0 \Rightarrow \lambda = -1$.
Substituting $\lambda = -1$ into the coordinates of $P$, we get $P = (3( -1)+6, 2(-1)+7, -2(-1)+7) = (3, 5, 9)$.
The length of the perpendicular $AP$ is the distance between $A(1, 2, 3)$ and $P(3, 5, 9)$:
$AP = \sqrt{(3-1)^2 + (5-2)^2 + (9-3)^2} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \text{ units}$.

(C) For two lines to intersect,the shortest distance between them must be $0$. The condition for intersection of two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ is given by the determinant: $\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right| = 0$.
Substituting the given values: $(x_1, y_1, z_1) = (k, -1, 1)$,$(a_1, b_1, c_1) = (2, 3, 4)$ and $(x_2, y_2, z_2) = (3, \frac{9}{2}, 0)$,$(a_2, b_2, c_2) = (1, 2, 1)$.
The determinant becomes: $\left|\begin{array}{ccc} 3-k & \frac{9}{2}-(-1) & 0-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right| = 0$.
$\Rightarrow \left|\begin{array}{ccc} 3-k & \frac{11}{2} & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right| = 0$.
Expanding the determinant: $(3-k)(3-8) - \frac{11}{2}(2-4) - 1(4-3) = 0$.
$\Rightarrow (3-k)(-5) - \frac{11}{2}(-2) - 1(1) = 0$.
$\Rightarrow -15 + 5k + 11 - 1 = 0$.
$\Rightarrow 5k - 5 = 0$.
$\Rightarrow 5k = 5$.
$\Rightarrow k = 1$.

(A) Let $\vec{n}$ be inclined at angles $\alpha, \beta, \gamma$ to $X, Y, Z$ axes respectively.
Given $\alpha = 45^{\circ}, \beta = 60^{\circ}$.
We know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the values: $\cos^2(45^{\circ}) + \cos^2(60^{\circ}) + \cos^2 \gamma = 1$.
$\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2 \gamma = 1$.
$\frac{1}{2} + \frac{1}{4} + \cos^2 \gamma = 1$.
$\frac{3}{4} + \cos^2 \gamma = 1 \implies \cos^2 \gamma = \frac{1}{4}$.
Since $\gamma$ is an acute angle,$\cos \gamma = \frac{1}{2}$,so $\gamma = 60^{\circ}$.
The normal vector is $\vec{n} = \cos \alpha \hat{i} + \cos \beta \hat{j} + \cos \gamma \hat{k} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} + \frac{1}{2} \hat{k}$.
Multiplying by $2$ to simplify the normal vector,we can take $\vec{n}' = \sqrt{2} \hat{i} + \hat{j} + \hat{k}$.
The equation of the plane passing through $(x_0, y_0, z_0) = (-\sqrt{2}, 1, 1)$ with normal $\vec{n}'$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
$\sqrt{2}(x - (-\sqrt{2})) + 1(y - 1) + 1(z - 1) = 0$.
$\sqrt{2}(x + \sqrt{2}) + y - 1 + z - 1 = 0$.
$\sqrt{2}x + 2 + y + z - 2 = 0$.
$\sqrt{2}x + y + z = 0$.

(A) The normal vector $\vec{n}$ to the plane is the vector from the origin $(0, 0, 0)$ to the foot of the perpendicular $(4, -2, 5)$.
Thus,$\vec{n} = (4 - 0)\hat{i} + (-2 - 0)\hat{j} + (5 - 0)\hat{k} = 4\hat{i} - 2\hat{j} + 5\hat{k}$.
The equation of a plane passing through a point $(x_0, y_0, z_0)$ with normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
Substituting the point $(4, -2, 5)$ and the normal vector $(4, -2, 5)$:
$4(x - 4) - 2(y - (-2)) + 5(z - 5) = 0$
$4(x - 4) - 2(y + 2) + 5(z - 5) = 0$
$4x - 16 - 2y - 4 + 5z - 25 = 0$
$4x - 2y + 5z - 45 = 0$
$4x - 2y + 5z = 45$.
Therefore,the correct option is $A$.

(C) The equation of a plane passing through the intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$.
Here,the planes are $2x - y - 4 = 0$ and $y + 2z - 4 = 0$.
So,the equation of the required plane is $(2x - y - 4) + \lambda(y + 2z - 4) = 0$ --- $(i)$.
Since the plane passes through the point $(2, 1, 0)$,we substitute $x = 2, y = 1, z = 0$ into equation $(i)$:
$(2(2) - 1 - 4) + \lambda(1 + 2(0) - 4) = 0$
$(4 - 1 - 4) + \lambda(1 - 4) = 0$
$-1 - 3\lambda = 0$
$-3\lambda = 1$
$\lambda = -\frac{1}{3}$.
Substituting $\lambda = -\frac{1}{3}$ back into equation $(i)$:
$(2x - y - 4) - \frac{1}{3}(y + 2z - 4) = 0$
Multiply the entire equation by $3$:
$3(2x - y - 4) - (y + 2z - 4) = 0$
$6x - 3y - 12 - y - 2z + 4 = 0$
$6x - 4y - 2z - 8 = 0$
Dividing by $2$:
$3x - 2y - z - 4 = 0$
$3x - 2y - z = 4$.

(D) The lines are given by $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$.
Comparing these with $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$,we get:
$(x_1, y_1, z_1) = (1, 2, 3)$ and $(x_2, y_2, z_2) = (2, 4, 5)$.
Direction ratios are $(a_1, b_1, c_1) = (2, 3, 4)$ and $(a_2, b_2, c_2) = (3, 4, 5)$.
The shortest distance $d$ is given by the formula:
$d = \frac{|\det(A)|}{\sqrt{(a_1b_2-a_2b_1)^2 + (b_1c_2-b_2c_1)^2 + (c_1a_2-c_2a_1)^2}}$
where $A = \begin{bmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix}$.
Calculating the determinant: $\det(A) = 1(15-16) - 2(10-12) + 2(8-9) = 1(-1) - 2(-2) + 2(-1) = -1 + 4 - 2 = 1$.
Now,calculate the denominator:
$a_1b_2-a_2b_1 = (2)(4)-(3)(3) = 8-9 = -1$.
$b_1c_2-b_2c_1 = (3)(5)-(4)(4) = 15-16 = -1$.
$c_1a_2-c_2a_1 = (4)(3)-(5)(2) = 12-10 = 2$.
Denominator $= \sqrt{(-1)^2 + (-1)^2 + (2)^2} = \sqrt{1+1+4} = \sqrt{6}$.
Therefore,$d = \frac{1}{\sqrt{6}}$ units.

(B) The perpendicular distance $d$ from a point $(x_1, y_1, z_1)$ to the plane $Ax+By+Cz+D=0$ is given by the formula $d = \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$.
Here,the point is the origin $(0, 0, 0)$ and the plane is $x-3y+4z-6=0$.
Substituting the values $A=1, B=-3, C=4, D=-6$ and $x_1=0, y_1=0, z_1=0$ into the formula:
$d = \frac{|1(0) - 3(0) + 4(0) - 6|}{\sqrt{1^2 + (-3)^2 + 4^2}}$
$d = \frac{|-6|}{\sqrt{1 + 9 + 16}}$
$d = \frac{6}{\sqrt{26}}$

(C) The equation of any plane passing through the intersection of the planes $P_1: x+y+z-1=0$ and $P_2: 2x+3y-z+4=0$ is given by $P_1 + \lambda P_2 = 0$.
$(x+y+z-1) + \lambda(2x+3y-z+4) = 0$
$(1+2\lambda)x + (1+3\lambda)y + (1-\lambda)z + (4\lambda-1) = 0$
Since this plane is parallel to the $Y$-axis,its normal vector $\vec{n} = (1+2\lambda, 1+3\lambda, 1-\lambda)$ must be perpendicular to the $Y$-axis unit vector $\hat{j} = (0, 1, 0)$.
Therefore,the coefficient of $y$ must be zero:
$1+3\lambda = 0 \Rightarrow \lambda = -\frac{1}{3}$.
Substituting $\lambda = -\frac{1}{3}$ into the equation:
$(x+y+z-1) - \frac{1}{3}(2x+3y-z+4) = 0$
$3(x+y+z-1) - (2x+3y-z+4) = 0$
$3x+3y+3z-3 - 2x-3y+z-4 = 0$
$x+4z-7 = 0$.
Checking the options,the point $(3, 2, 1)$ satisfies the equation: $3 + 4(1) - 7 = 3+4-7 = 0$.

(C) The equation of a plane passing through $(1, 1, 1)$ is given by $a(x-1) + b(y-1) + c(z-1) = 0$.
Since the plane is parallel to the lines with direction ratios $(1, 0, -1)$ and $(-1, 1, 0)$,the normal vector $(a, b, c)$ must be perpendicular to both direction vectors.
Thus,$a(1) + b(0) + c(-1) = 0 \Rightarrow a - c = 0$ and $a(-1) + b(1) + c(0) = 0 \Rightarrow -a + b = 0$.
This implies $a = c$ and $a = b$,so $a = b = c$.
Substituting $a=b=c$ into the plane equation,we get $a(x-1) + a(y-1) + a(z-1) = 0$,which simplifies to $x + y + z = 3$.
Dividing by $3$,we get the intercept form $\frac{x}{3} + \frac{y}{3} + \frac{z}{3} = 1$.
Thus,the coordinates of $A, B, C$ are $(3, 0, 0), (0, 3, 0),$ and $(0, 0, 3)$ respectively.
The volume of the tetrahedron $OABC$ is given by $V = \frac{1}{6} |x_A y_B z_C| = \frac{1}{6} |3 \times 3 \times 3| = \frac{27}{6} = \frac{9}{2}$ cubic units.

(B) Let the normal vector to the plane be $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$. Since the normal makes equal acute angles with the coordinate axes,the direction cosines are equal,i.e.,$l = m = n$.
Thus,the normal vector is $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.
The equation of a plane passing through a point $\vec{r}_0 = (-1, 1, 2)$ with normal $\vec{n}$ is given by $(\vec{r} - \vec{r}_0) \cdot \vec{n} = 0$.
Substituting the values:
$(x - (-1))\hat{i} + (y - 1)\hat{j} + (z - 2)\hat{k} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$
$(x + 1) + (y - 1) + (z - 2) = 0$
$x + y + z - 2 = 0$.
Therefore,the correct option is $(B)$.

(C) First,find the equation of the plane containing the lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$. The normal vector $\vec{n_1}$ to this plane is given by the cross product of the direction vectors $\vec{v_1} = (3, 4, 2)$ and $\vec{v_2} = (4, 2, 3)$.
$\vec{n_1} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 2 \\ 4 & 2 & 3 \end{vmatrix} = \hat{i}(12-4) - \hat{j}(9-8) + \hat{k}(6-16) = 8\hat{i} - \hat{j} - 10\hat{k}$.
The equation of this plane is $8x - y - 10z = 0$.
Let the required plane be $ax + by + cz = 0$. Since it contains the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$,its normal vector $\vec{n_2} = (a, b, c)$ must be perpendicular to the line's direction vector $\vec{v_3} = (2, 3, 4)$. Thus,$2a + 3b + 4c = 0$.
Since the required plane is perpendicular to the plane $8x - y - 10z = 0$,their normal vectors are perpendicular: $(a, b, c) \cdot (8, -1, -10) = 0$,so $8a - b - 10c = 0$.
Solving these two equations for $a, b, c$ using cross product: $\vec{n_2} = (3, 4, 2) \times (8, -1, -10) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 8 & -1 & -10 \end{vmatrix} = \hat{i}(-30+4) - \hat{j}(-20-32) + \hat{k}(-2-24) = -26\hat{i} + 52\hat{j} - 26\hat{k}$.
Dividing by $-26$,we get the normal vector $(1, -2, 1)$.
The equation of the plane is $1(x) - 2(y) + 1(z) = 0$,which is $x - 2y + z = 0$.

(C) The normal vector $\vec{n}$ to the plane is the cross product of the direction vectors of the two lines: $\vec{n} = (2, 0, -2) \times (-2, 2, 0)$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -2 \\ -2 & 2 & 0 \end{vmatrix} = \hat{i}(0 - (-4)) - \hat{j}(0 - 4) + \hat{k}(4 - 0) = 4\hat{i} + 4\hat{j} + 4\hat{k}$.
We can take the normal vector as $\vec{n} = (1, 1, 1)$.
The equation of the plane passing through $(2, 2, 2)$ is $1(x-2) + 1(y-2) + 1(z-2) = 0$,which simplifies to $x + y + z = 6$.
The intercepts on the axes are $A(6, 0, 0)$,$B(0, 6, 0)$,and $C(0, 0, 6)$.
The volume of the tetrahedron $OABC$ is given by $V = \frac{1}{6} |x_{int} \cdot y_{int} \cdot z_{int}|$.
$V = \frac{1}{6} |6 \times 6 \times 6| = \frac{216}{6} = 36 \text{ cubic units}$.

(A) The equation of a plane passing through $(1, -1, 2)$ is given by $a(x - 1) + b(y + 1) + c(z - 2) = 0$.
Since this plane is perpendicular to the planes $x + 2y - 2z = 4$ and $3x + 2y + z = 6$,its normal vector $(a, b, c)$ must be perpendicular to the normal vectors of the given planes,which are $\vec{n_1} = (1, 2, -2)$ and $\vec{n_2} = (3, 2, 1)$.
Thus,we have the equations:
$a + 2b - 2c = 0$
$3a + 2b + c = 0$
Using the cross product to find the direction ratios $(a, b, c)$:
$a = (2)(1) - (-2)(2) = 2 + 4 = 6$
$b = (-2)(3) - (1)(1) = -6 - 1 = -7$
$c = (1)(2) - (2)(3) = 2 - 6 = -4$
So,the direction ratios are $(6, -7, -4)$.
Substituting these into the plane equation:
$6(x - 1) - 7(y + 1) - 4(z - 2) = 0$
$6x - 6 - 7y - 7 - 4z + 8 = 0$
$6x - 7y - 4z - 5 = 0$.

(D) The normal vector to the plane $x - y + 2z - 2 = 0$ is $\vec{n} = (1, -1, 2)$.
Let $PQ$ be the line passing through $P(2, 4, -1)$ and perpendicular to the plane. The equation of line $PQ$ is:
$\frac{x - 2}{1} = \frac{y - 4}{-1} = \frac{z + 1}{2} = \lambda$
So,any point on the line is $M(\lambda + 2, 4 - \lambda, 2\lambda - 1)$.
Since $M$ lies on the plane,it satisfies the plane equation:
$(\lambda + 2) - (4 - \lambda) + 2(2\lambda - 1) - 2 = 0$
$\lambda + 2 - 4 + \lambda + 4\lambda - 2 - 2 = 0$
$6\lambda - 6 = 0 \Rightarrow \lambda = 1$.
Substituting $\lambda = 1$,we get the coordinates of $M$ as $(3, 3, 1)$.
Since $M$ is the midpoint of $PQ$ and $Q = (a, b, c)$:
$\frac{2 + a}{2} = 3 \Rightarrow a = 4$
$\frac{4 + b}{2} = 3 \Rightarrow b = 2$
$\frac{-1 + c}{2} = 1 \Rightarrow c = 3$
Thus,$a + b + c = 4 + 2 + 3 = 9$.