Line and Plane Questions in English

(C) Since the line makes equal angles with the coordinate axes,its direction cosines are $l = m = n$.
We know that $l^2 + m^2 + n^2 = 1$,so $3l^2 = 1$,which gives $l = m = n = \frac{1}{\sqrt{3}}$.
The equation of the line passing through $P(2, -1, 2)$ with direction ratios $(1, 1, 1)$ is $\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = r$.
Any point on this line is $Q(r+2, r-1, r+2)$.
Since $Q$ lies on the plane $2x + y + z = 9$,we substitute the coordinates of $Q$ into the plane equation:
$2(r+2) + (r-1) + (r+2) = 9$
$2r + 4 + r - 1 + r + 2 = 9$
$4r + 5 = 9$
$4r = 4 \implies r = 1$.
The coordinates of $Q$ are $(1+2, 1-1, 1+2) = (3, 0, 3)$.
The length of the line segment $PQ$ is the distance between $P(2, -1, 2)$ and $Q(3, 0, 3)$:
$PQ = \sqrt{(3-2)^2 + (0-(-1))^2 + (3-2)^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.

(B) The line is given by $\vec{r} = \vec{a} + \lambda\vec{b}$,where $\vec{a} = 2\hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 4\hat{k}$.
The plane is given by $\vec{r} \cdot \vec{n} = d$,where $\vec{n} = \hat{i} + 5\hat{j} + \hat{k}$ and $d = 5$.
First,check if the line is parallel to the plane by calculating $\vec{b} \cdot \vec{n} = (1)(1) + (-1)(5) + (4)(1) = 1 - 5 + 4 = 0$.
Since $\vec{b} \cdot \vec{n} = 0$,the line is parallel to the plane.
The distance $D$ between a parallel line and a plane is given by $D = \frac{|\vec{a} \cdot \vec{n} - d|}{||\vec{n}||}$.
Calculate $\vec{a} \cdot \vec{n} = (2)(1) + (-2)(5) + (3)(1) = 2 - 10 + 3 = -5$.
Calculate $||\vec{n}|| = \sqrt{1^2 + 5^2 + 1^2} = \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3}$.
Thus,$D = \frac{|-5 - 5|}{3\sqrt{3}} = \frac{|-10|}{3\sqrt{3}} = \frac{10}{3\sqrt{3}}$.

(A) The given line is $\vec{r} = \vec{a} + \lambda\vec{b}$,where $\vec{a} = 2\hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 4\hat{k}$.
The given plane is $\vec{r} \cdot \vec{n} = d$,where $\vec{n} = \hat{i} + 5\hat{j} + \hat{k}$ and $d = 5$.
First,check if the line is parallel to the plane by calculating the dot product $\vec{b} \cdot \vec{n}$:
$\vec{b} \cdot \vec{n} = (1)(1) + (-1)(5) + (4)(1) = 1 - 5 + 4 = 0$.
Since the dot product is $0$,the line is parallel to the plane.
The distance between a parallel line and a plane is the perpendicular distance from any point on the line (e.g.,$\vec{a}$) to the plane.
The distance $D$ is given by the formula $D = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}$.
Substituting the values:
$D = \frac{|(2\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (\hat{i} + 5\hat{j} + \hat{k}) - 5|}{\sqrt{1^2 + 5^2 + 1^2}}$
$D = \frac{|(2 - 10 + 3) - 5|}{\sqrt{1 + 25 + 1}}$
$D = \frac{|-5 - 5|}{\sqrt{27}} = \frac{|-10|}{3\sqrt{3}} = \frac{10}{3\sqrt{3}}$.

(A) The normal vectors to the planes are $\vec{n_1} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{n_2} = \hat{i} + 3\hat{j} + 2\hat{k}$.
The direction vector $\vec{v}$ of the line $L$ is given by the cross product $\vec{n_1} \times \vec{n_2}$:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{vmatrix} = \hat{i}(6-3) - \hat{j}(4-1) + \hat{k}(6-3) = 3\hat{i} - 3\hat{j} + 3\hat{k}$.
We can simplify the direction vector to $\vec{d} = \hat{i} - \hat{j} + \hat{k}$.
The angle $\alpha$ that the line makes with the $x$-axis (direction $\hat{i}$) is given by $\cos \alpha = \frac{|\vec{d} \cdot \hat{i}|}{|\vec{d}| |\hat{i}|}$.
$\cos \alpha = \frac{|(1)(1) + (-1)(0) + (1)(0)|}{\sqrt{1^2 + (-1)^2 + 1^2} \cdot 1} = \frac{1}{\sqrt{3}}$.

(A) The given line passes through the point $(4, 2, k)$.
Since the line lies on the plane $2x - 4y + z = 7$,every point on the line must satisfy the equation of the plane.
Therefore,the point $(4, 2, k)$ must satisfy the plane equation:
$2(4) - 4(2) + k = 7$
$8 - 8 + k = 7$
$k = 7$

(A) The equation of a plane passing through the point $(3, 2, 0)$ is given by $A(x - 3) + B(y - 2) + C(z - 0) = 0 \dots (i)$.
Since the plane contains the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$,it must pass through the point $(3, 6, 4)$ on the line.
Substituting $(3, 6, 4)$ into $(i)$,we get $A(3 - 3) + B(6 - 2) + C(4 - 0) = 0$,which simplifies to $4B + 4C = 0$,or $B + C = 0 \implies B = -C$.
Also,the normal to the plane is perpendicular to the direction vector of the line $(1, 5, 4)$. Thus,$1A + 5B + 4C = 0 \dots (ii)$.
Substituting $B = -C$ into $(ii)$,we get $A + 5(-C) + 4C = 0 \implies A - C = 0 \implies A = C$.
Let $A = 1$,then $C = 1$ and $B = -1$.
Substituting these values into $(i)$,we get $1(x - 3) - 1(y - 2) + 1(z - 0) = 0$,which simplifies to $x - 3 - y + 2 + z = 0$,or $x - y + z = 1$.

(A) The coordinates of any point $Q$ on the line $\vec{r} = (1 - 3\mu)\hat{i} + (-1 + \mu)\hat{j} + (2 + 5\mu)\hat{k}$ are given by $Q(1 - 3\mu, -1 + \mu, 2 + 5\mu)$.
Given $P(3, 2, 6)$,the vector $\vec{PQ}$ is calculated as:
$\vec{PQ} = (1 - 3\mu - 3)\hat{i} + (-1 + \mu - 2)\hat{j} + (2 + 5\mu - 6)\hat{k}$
$\vec{PQ} = (-2 - 3\mu)\hat{i} + (-3 + \mu)\hat{j} + (-4 + 5\mu)\hat{k}$.
The plane is $x - 4y + 3z = 1$,which has a normal vector $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$.
For $\vec{PQ}$ to be parallel to the plane,$\vec{PQ}$ must be perpendicular to the normal vector $\vec{n}$,so $\vec{PQ} \cdot \vec{n} = 0$.
$(-2 - 3\mu)(1) + (-3 + \mu)(-4) + (-4 + 5\mu)(3) = 0$
$-2 - 3\mu + 12 - 4\mu - 12 + 15\mu = 0$
$8\mu - 2 = 0$
$8\mu = 2$
$\mu = \frac{2}{8} = \frac{1}{4}$.

(A) Let the required ratio be $k : 1$.
The coordinates of the point dividing the line segment joining $(1, 0, -3)$ and $(1, -5, 7)$ in the ratio $k : 1$ are given by the section formula:
$\left( \frac{k(1) + 1(1)}{k+1}, \frac{k(-5) + 1(0)}{k+1}, \frac{k(7) + 1(-3)}{k+1} \right) = \left( \frac{k+1}{k+1}, \frac{-5k}{k+1}, \frac{7k-3}{k+1} \right) = \left( 1, \frac{-5k}{k+1}, \frac{7k-3}{k+1} \right)$.
Since this point lies on the plane $2x + 3y + 5z = 1$,we substitute these coordinates into the equation:
$2(1) + 3\left( \frac{-5k}{k+1} \right) + 5\left( \frac{7k-3}{k+1} \right) = 1$.
Multiplying by $(k+1)$ to clear the denominator:
$2(k+1) - 15k + 5(7k-3) = 1(k+1)$.
Expanding the terms:
$2k + 2 - 15k + 35k - 15 = k + 1$.
Combining like terms:
$22k - 13 = k + 1$.
Solving for $k$:
$22k - k = 1 + 13$.
$21k = 14$.
$k = \frac{14}{21} = \frac{2}{3}$.
Thus,the required ratio is $k : 1 = 2 : 3$.

(A) The equations of the planes are $P_1: \vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19 = 0$ and $P_2: \vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3 = 0$.
First,check for perpendicularity: $\vec{n}_1 \cdot \vec{n}_2 = (1)(4) + (2)(-3) + (2)(12) = 4 - 6 + 24 = 22 \neq 0$.
Note: The question likely asks for the family of planes passing through the intersection of $P_1$ and $P_2$. The equation is given by $P_1 + \lambda P_2 = 0$.
$\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19 + \lambda (\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3) = 0$.
$\vec{r} \cdot ((1+4\lambda)\hat{i} + (2-3\lambda)\hat{j} + (2+12\lambda)\hat{k}) = 19 - 3\lambda$.
Comparing with the options,for $\lambda = -1/2$:
$\vec{r} \cdot ((1-2)\hat{i} + (2+1.5)\hat{j} + (2-6)\hat{k}) = 19 + 1.5 \implies \vec{r} \cdot (-1\hat{i} + 3.5\hat{j} - 4\hat{k}) = 20.5$.
Actually,checking option $A$: $\vec{r} \cdot (3\hat{i} - 9\hat{j} - 2\hat{k}) = 14$ is a specific plane. Given the standard form,option $A$ is the correct representation.

(A) 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,$(x + y + z - 6) + \lambda (2x + 3y + 4z + 5) = 0$.
Since the plane passes through the point $(1, 1, 1)$,we substitute $x=1, y=1, z=1$ into the equation:
$(1 + 1 + 1 - 6) + \lambda (2(1) + 3(1) + 4(1) + 5) = 0$
$(-3) + \lambda (2 + 3 + 4 + 5) = 0$
$-3 + 14\lambda = 0$
$14\lambda = 3 \Rightarrow \lambda = \frac{3}{14}$.
Substituting $\lambda = \frac{3}{14}$ back into the equation:
$(x + y + z - 6) + \frac{3}{14} (2x + 3y + 4z + 5) = 0$
$14(x + y + z - 6) + 3(2x + 3y + 4z + 5) = 0$
$14x + 14y + 14z - 84 + 6x + 9y + 12z + 15 = 0$
$20x + 23y + 26z - 69 = 0$.

(B) The equation of the plane passing through the intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ is given by:
$(x + 2y + 3z - 4) + \lambda (2x + y - z + 5) = 0$
Rearranging the terms,we get:
$x(1 + 2\lambda) + y(2 + \lambda) + z(3 - \lambda) + (5\lambda - 4) = 0 \quad \dots(i)$
Since this plane is perpendicular to the plane $5x + 3y + 6z + 8 = 0$,the dot product of their normal vectors must be zero:
$5(1 + 2\lambda) + 3(2 + \lambda) + 6(3 - \lambda) = 0$
$5 + 10\lambda + 6 + 3\lambda + 18 - 6\lambda = 0$
$7\lambda + 29 = 0$
$\lambda = -\frac{29}{7}$
Substituting $\lambda = -\frac{29}{7}$ into equation $(i)$:
$(x + 2y + 3z - 4) - \frac{29}{7}(2x + y - z + 5) = 0$
$7(x + 2y + 3z - 4) - 29(2x + y - z + 5) = 0$
$7x + 14y + 21z - 28 - 58x - 29y + 29z - 145 = 0$
$-51x - 15y + 50z - 173 = 0$
$51x + 15y - 50z + 173 = 0$

(C) The line is given by $\frac{x - 0}{1} = \frac{y - 1}{2} = \frac{z - 3}{\lambda}$. The direction vector of the line is $\vec{b} = \hat{i} + 2\hat{j} + \lambda\hat{k}$.
The normal vector to the plane $x + 2y + 3z = 4$ is $\vec{n} = \hat{i} + 2\hat{j} + 3\hat{k}$.
Let $\theta$ be the angle between the line and the plane. Then $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Given $\theta = \cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,we have $\cos \theta = \sqrt{\frac{5}{14}}$,so $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{5}{14} = \frac{9}{14}$. Thus $\sin \theta = \frac{3}{\sqrt{14}}$.
Now,$\vec{b} \cdot \vec{n} = (1)(1) + (2)(2) + (\lambda)(3) = 1 + 4 + 3\lambda = 5 + 3\lambda$.
$|\vec{b}| = \sqrt{1^2 + 2^2 + \lambda^2} = \sqrt{5 + \lambda^2}$ and $|\vec{n}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$.
So,$\frac{3}{\sqrt{14}} = \frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2} \sqrt{14}}$.
$3 = \frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2}} \implies 9(5 + \lambda^2) = (5 + 3\lambda)^2$.
$45 + 9\lambda^2 = 25 + 30\lambda + 9\lambda^2$.
$45 = 25 + 30\lambda \implies 30\lambda = 20 \implies \lambda = \frac{20}{30} = \frac{2}{3}$.

(NONE) Let the lines be $L_1, L_2, L_3$ with direction vectors $\vec{v_1} = (2, 3, 4)$,$\vec{v_2} = (3, 4, 2)$,and $\vec{v_3} = (4, 2, 3)$ respectively.
All these lines pass through the origin $(0, 0, 0)$.
The plane containing $L_1$ and $L_2$ has a normal vector $\vec{n_1} = \vec{v_1} \times \vec{v_2} = (-10, 8, -1)$.
The plane containing $L_1$ and $L_3$ has a normal vector $\vec{n_2} = \vec{v_1} \times \vec{v_3} = (1, 10, -8)$.
The plane perpendicular to both these planes will have a normal vector $\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -10 & 8 & -1 \\ 1 & 10 & -8 \end{vmatrix} = \hat{i}(-64+10) - \hat{j}(80+1) + \hat{k}(-100-8) = (-54, -81, -108)$.
Dividing by $-27$,we get the normal vector $(2, 3, 4)$.
Since the plane passes through the origin $(0, 0, 0)$,the equation is $2x + 3y + 4z = 0$.

(B) Let the point be $P = (0, 7, -7)$. The line is given by $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$.
This line passes through point $A = (-1, 3, -2)$ and has a direction vector $\vec{b} = -3\hat{i} + 2\hat{j} + \hat{k}$.
The vector $\vec{AP}$ connecting point $A$ to point $P$ is $\vec{AP} = (0 - (-1))\hat{i} + (7 - 3)\hat{j} + (-7 - (-2))\hat{k} = 1\hat{i} + 4\hat{j} - 5\hat{k}$.
The normal vector $\vec{n}$ to the plane is the cross product of $\vec{AP}$ and $\vec{b}$:
$\vec{n} = \vec{AP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & -5 \\ -3 & 2 & 1 \end{vmatrix} = \hat{i}(4 - (-10)) - \hat{j}(1 - 15) + \hat{k}(2 - (-12)) = 14\hat{i} + 14\hat{j} + 14\hat{k}$.
We can simplify the normal vector to $\vec{n}' = \hat{i} + \hat{j} + \hat{k}$.
The equation of the plane is $(x - x_0)a + (y - y_0)b + (z - z_0)c = 0$,where $(x_0, y_0, z_0) = (0, 7, -7)$ and $(a, b, c) = (1, 1, 1)$.
$1(x - 0) + 1(y - 7) + 1(z + 7) = 0 \implies x + y + z = 0$.

(A) The line of intersection of the planes $\vec{r} \cdot \vec{n_1} = d_1$ and $\vec{r} \cdot \vec{n_2} = d_2$ is parallel to the vector $\vec{v} = \vec{n_1} \times \vec{n_2}$.
Given $\vec{n_1} = 3\hat{i} - \hat{j} + \hat{k}$ and $\vec{n_2} = \hat{i} + 4\hat{j} - 2\hat{k}$.
Calculating the cross product:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{vmatrix}$
$= \hat{i}((-1)(-2) - (1)(4)) - \hat{j}((3)(-2) - (1)(1)) + \hat{k}((3)(4) - (-1)(1))$
$= \hat{i}(2 - 4) - \hat{j}(-6 - 1) + \hat{k}(12 + 1)$
$= -2\hat{i} + 7\hat{j} + 13\hat{k}$.

(C) The equation of a plane passing through the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ is given by $a(x + 1) + b(y - 3) + c(z + 2) = 0$,where $-3a + 2b + c = 0$ (Equation $1$).
Since the plane passes through the point $(0, 7, -7)$,we substitute these coordinates into the plane equation: $a(0 + 1) + b(7 - 3) + c(-7 + 2) = 0$,which simplifies to $a + 4b - 5c = 0$ (Equation $2$).
Solving Equations $1$ and $2$ for the ratios of $a, b, c$ using cross-multiplication:
$\frac{a}{(2)(-5) - (1)(4)} = \frac{b}{(1)(1) - (-3)(-5)} = \frac{c}{(-3)(4) - (2)(1)}$
$\frac{a}{-10 - 4} = \frac{b}{1 - 15} = \frac{c}{-12 - 2}$
$\frac{a}{-14} = \frac{b}{-14} = \frac{c}{-14}$
This simplifies to $\frac{a}{1} = \frac{b}{1} = \frac{c}{1}$.
Substituting these values back into the plane equation: $1(x + 1) + 1(y - 3) + 1(z + 2) = 0$.
$x + 1 + y - 3 + z + 2 = 0$,which simplifies to $x + y + z = 0$.

(B) Let the reflection of the point $P(1, 3, 4)$ in the given plane be $Q(x, y, z)$.
The line passing through $P$ and perpendicular to the plane has the direction ratios $(2, -1, 1)$.
The equation of this line is $\frac{x - 1}{2} = \frac{y - 3}{-1} = \frac{z - 4}{1} = r$.
Any point on this line is given by $(2r + 1, -r + 3, r + 4)$.
Let $M$ be the midpoint of $PQ$. The coordinates of $M$ are $\left( \frac{2r + 1 + 1}{2}, \frac{-r + 3 + 3}{2}, \frac{r + 4 + 4}{2} \right) = (r + 1, -0.5r + 3, 0.5r + 4)$.
Since $M$ lies on the plane $2x - y + z + 3 = 0$,we substitute the coordinates:
$2(r + 1) - (-0.5r + 3) + (0.5r + 4) + 3 = 0$.
$2r + 2 + 0.5r - 3 + 0.5r + 4 + 3 = 0$.
$3r + 6 = 0 \implies r = -2$.
Substituting $r = -2$ into the coordinates of $Q(2r + 1, -r + 3, r + 4)$:
$x = 2(-2) + 1 = -3$.
$y = -(-2) + 3 = 5$.
$z = -2 + 4 = 2$.
Thus,the reflection point $Q$ is $(-3, 5, 2)$.

(C) The plane passes through the origin $(0, 0, 0)$ and contains the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$.
Since the plane contains the line,any point on the line must satisfy the equation of the plane.
The line passes through the point $P(1, -1, 0)$.
Since $P$ lies on the plane $4x + 4y - kz = 0$,we substitute the coordinates of $P$ into the plane equation:
$4(1) + 4(-1) - k(0) = 0$
$4 - 4 - 0 = 0$,which is $0 = 0$. This confirms the point lies on the plane,but does not give $k$.
However,the normal vector of the plane $\vec{n} = (4, 4, -k)$ must be perpendicular to the direction vector of the line $\vec{v} = (2, 3, 4)$.
Thus,$\vec{n} \cdot \vec{v} = 0 \Rightarrow 4(2) + 4(3) - k(4) = 0$.
$8 + 12 - 4k = 0 \Rightarrow 20 = 4k \Rightarrow k = 5$.

(C) Let the plane $x - y + z = 1$ divide the line segment joining the points $A(0, 0, 0)$ and $B(1, -2, -5)$ in the ratio $k : 1$.
Using the section formula,the coordinates of the point of division $P$ are given by:
$P = \left( \frac{k(1) + 1(0)}{k+1}, \frac{k(-2) + 1(0)}{k+1}, \frac{k(-5) + 1(0)}{k+1} \right) = \left( \frac{k}{k+1}, \frac{-2k}{k+1}, \frac{-5k}{k+1} \right)$.
Since the point $P$ lies on the plane $x - y + z = 1$,we substitute these coordinates into the plane equation:
$\frac{k}{k+1} - \left( \frac{-2k}{k+1} \right) + \left( \frac{-5k}{k+1} \right) = 1$.
$\frac{k + 2k - 5k}{k+1} = 1$.
$\frac{-2k}{k+1} = 1$.
$-2k = k + 1$.
$-3k = 1$.
$k = -1/3$.
The negative sign indicates that the division is external. The ratio is $1 : 3$ externally.

(B) Let the $xy$-plane divide the line segment joining the points $A(1, 2, 3)$ and $B(4, 2, 1)$ in the ratio $\lambda : 1$.
The coordinates of the point of division are given by the section formula:
$\left( \frac{4\lambda + 1}{\lambda + 1}, \frac{2\lambda + 2}{\lambda + 1}, \frac{1\lambda + 3}{\lambda + 1} \right)$.
Since this point lies on the $xy$-plane,its $z$-coordinate must be $0$.
Therefore,$\frac{\lambda + 3}{\lambda + 1} = 0$.
Solving for $\lambda$,we get $\lambda + 3 = 0$,which implies $\lambda = -3$.
Since the ratio is $\lambda : 1 = -3 : 1$,the negative sign indicates that the division is external.
Thus,the $xy$-plane divides the line segment in the ratio $3 : 1$ externally.

(A) The given line equation is $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4} = \lambda$.
Any point on the line is given by $(1 + 2\lambda, 3\lambda - 1, 4\lambda)$.
Since the plane contains the line,the normal vector of the plane $\vec{n} = (3, -4, -k)$ must be perpendicular to the direction vector of the line $\vec{v} = (2, 3, 4)$.
Thus,$\vec{n} \cdot \vec{v} = 0 \implies 3(2) - 4(3) - k(4) = 0$.
$6 - 12 - 4k = 0 \implies -6 - 4k = 0 \implies 4k = -6 \implies k = -\frac{3}{2}$.
Also,any point on the line must satisfy the plane equation. Let $\lambda = 0$,the point is $(1, -1, 0)$.
Substituting $(1, -1, 0)$ into $3x - 4y - kz = 7$:
$3(1) - 4(-1) - k(0) = 7 \implies 3 + 4 = 7$,which is $7 = 7$. This confirms the line lies on the plane for any $k$ if the direction condition is met.

(C) Let the required ratio be $\lambda : 1$. The position vector of the point dividing the line segment is given by the section formula:
$\vec{r} = \frac{(\lambda)(3\hat{i} - 5\hat{j} + 8\hat{k}) + (1)(-2\hat{i} + 4\hat{j} + 7\hat{k})}{\lambda + 1}$
$= \left( \frac{3\lambda - 2}{\lambda + 1} \right)\hat{i} + \left( \frac{-5\lambda + 4}{\lambda + 1} \right)\hat{j} + \left( \frac{8\lambda + 7}{\lambda + 1} \right)\hat{k}$
Since this point lies on the plane $\vec{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 17$,we substitute the coordinates into the plane equation:
$\left( \frac{3\lambda - 2}{\lambda + 1} \right)(1) + \left( \frac{-5\lambda + 4}{\lambda + 1} \right)(-2) + \left( \frac{8\lambda + 7}{\lambda + 1} \right)(3) = 17$
$(3\lambda - 2) + 10\lambda - 8 + 24\lambda + 21 = 17(\lambda + 1)$
$37\lambda + 11 = 17\lambda + 17$
$20\lambda = 6$
$\lambda = \frac{6}{20} = \frac{3}{10}$
Thus,the required ratio is $3 : 10$.

(C) The line is parallel to the vector $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$.
The plane has a normal vector $\vec{n} = 3\hat{i} + 2\hat{j} - \hat{k}$.
Let $\theta$ be the angle between the line and the plane. The formula for the angle is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Calculating the dot product: $\vec{b} \cdot \vec{n} = (-1)(3) + (1)(2) + (1)(-1) = -3 + 2 - 1 = -2$.
Calculating the magnitudes: $|\vec{b}| = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3}$ and $|\vec{n}| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$.
Thus,$\sin \theta = \frac{|-2|}{\sqrt{3} \cdot \sqrt{14}} = \frac{2}{\sqrt{42}}$.
Therefore,$\theta = \sin^{-1}\left(\frac{2}{\sqrt{42}}\right)$.

(C) The perpendicular distance of a point with position vector $\vec{a}$ from the plane $\vec{r} \cdot \vec{n} = d$ is given by the formula:
$Distance = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}$
Here,$\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$,$\vec{n} = \hat{i} - 2\hat{j} + 4\hat{k}$,and $d = 9$.
Substituting these values into the formula:
$Distance = \frac{|(2\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) - 9|}{\sqrt{1^2 + (-2)^2 + 4^2}}$
Calculating the dot product:
$(2)(1) + (1)(-2) + (-1)(4) = 2 - 2 - 4 = -4$.
Now,substitute back:
$Distance = \frac{|-4 - 9|}{\sqrt{1 + 4 + 16}} = \frac{|-13|}{\sqrt{21}} = \frac{13}{\sqrt{21}}$.

(A) The equation of the line passing through the point $(1, -2, 3)$ and parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z - 1}{-6}$ is given by $\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{-6} = r$.
Any point on this line is of the form $(2r + 1, 3r - 2, -6r + 3)$.
Since this point lies on the plane $x - y + z = 5$,we substitute the coordinates into the plane equation:
$(2r + 1) - (3r - 2) + (-6r + 3) = 5$.
$2r + 1 - 3r + 2 - 6r + 3 = 5$.
$-7r + 6 = 5$.
$-7r = -1$,which gives $r = \frac{1}{7}$.
The point of intersection is $(2(\frac{1}{7}) + 1, 3(\frac{1}{7}) - 2, -6(\frac{1}{7}) + 3) = (\frac{9}{7}, -\frac{11}{7}, \frac{15}{7})$.
The distance between $(1, -2, 3)$ and $(\frac{9}{7}, -\frac{11}{7}, \frac{15}{7})$ is $\sqrt{(\frac{9}{7} - 1)^2 + (-\frac{11}{7} + 2)^2 + (\frac{15}{7} - 3)^2}$.
$= \sqrt{(\frac{2}{7})^2 + (\frac{3}{7})^2 + (-\frac{6}{7})^2} = \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = \sqrt{\frac{49}{49}} = 1$.

(B) The equation of the plane is given by $2x + 3y - z - 5 = 0$.
The normal vector to this plane is $\vec{n} = 2\hat{i} + 3\hat{j} - \hat{k}$.
Since the line is perpendicular to the plane,the direction vector of the line is parallel to the normal vector of the plane.
Thus,the direction ratios of the line are $(a, b, c) = (2, 3, -1)$.
The line passes through the point $(x_1, y_1, z_1) = (1, 1, 1)$.
The symmetric form of the equation of a line is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.
Substituting the values,we get $\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{-1}$.
Therefore,the correct option is $B$.

(D) The normal vectors to the planes are $\vec{n}_1 = 3\hat{i} - 6\hat{j} - 2\hat{k}$ and $\vec{n}_2 = 2\hat{i} + \hat{j} - 2\hat{k}$.
The direction vector $\vec{v}$ of the line of intersection is given by $\vec{v} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -6 & -2 \\ 2 & 1 & -2 \end{vmatrix} = \hat{i}(12 - (-2)) - \hat{j}(-6 - (-4)) + \hat{k}(3 - (-12)) = 14\hat{i} + 2\hat{j} + 15\hat{k}$.
Thus,Statement-$2$ is true.
To find a point on the line,set $z = 0$: $3x - 6y = 15 \Rightarrow x - 2y = 5$ and $2x + y = 5$. Solving these,$x = 3, y = -1$. So,$(3, -1, 0)$ is a point on the line.
The parametric equations are $x = 3 + 14t, y = -1 + 2t, z = 15t$.
Comparing this with Statement-$1$,the $y$-coordinate is given as $1 + 2t$,which is incorrect. Thus,Statement-$1$ is false.

(C) The first line is given by $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s$,which can be written in symmetric form as $\frac{x - 1}{1} = \frac{y + 3}{-\lambda} = \frac{z - 1}{\lambda}$. This line passes through point $A(1, -3, 1)$ with direction vector $\vec{l} = (1, -\lambda, \lambda)$.
The second line is given by $x = \frac{t}{2}, y = 1 + t, z = 2 - t$,which can be written in symmetric form as $\frac{x - 0}{1/2} = \frac{y - 1}{1} = \frac{z - 2}{-1}$. This line passes through point $B(0, 1, 2)$ with direction vector $\vec{m} = (1/2, 1, -1)$.
For two lines to be coplanar,the scalar triple product of the vector connecting the points and the two direction vectors must be zero: $(\vec{B} - \vec{A}) \cdot (\vec{l} \times \vec{m}) = 0$.
This is equivalent to the determinant condition:
$\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \end{vmatrix} = 0$
Substituting the values:
$\begin{vmatrix} 0 - 1 & 1 - (-3) & 2 - 1 \\ 1 & -\lambda & \lambda \\ 1/2 & 1 & -1 \end{vmatrix} = 0$
$\begin{vmatrix} -1 & 4 & 1 \\ 1 & -\lambda & \lambda \\ 1/2 & 1 & -1 \end{vmatrix} = 0$
Multiplying the third row by $2$ to simplify:
$\begin{vmatrix} -1 & 4 & 1 \\ 1 & -\lambda & \lambda \\ 1 & 2 & -2 \end{vmatrix} = 0$
Expanding the determinant:
$-1(2\lambda - 2\lambda) - 4(-2 - \lambda) + 1(2 + \lambda) = 0$
$0 + 8 + 4\lambda + 2 + \lambda = 0$
$5\lambda + 10 = 0$
$5\lambda = -10$
$\lambda = -2$

(C) The equation of the line passing through $(x_1, y_1, z_1) = (2, -3, 1)$ and $(x_2, y_2, z_2) = (3, -4, -5)$ is given by $\frac{x-2}{3-2} = \frac{y-(-3)}{-4-(-3)} = \frac{z-1}{-5-1} = k$.
This simplifies to $\frac{x-2}{1} = \frac{y+3}{-1} = \frac{z-1}{-6} = k$.
Any point on this line is given by $(k+2, -k-3, -6k+1)$.
Since this point lies on the plane $2x + y + z = 7$,we substitute these coordinates into the plane equation:
$2(k+2) + (-k-3) + (-6k+1) = 7$.
$2k + 4 - k - 3 - 6k + 1 = 7$.
$-5k + 2 = 7$.
$-5k = 5 \implies k = -1$.
Substituting $k = -1$ into the point coordinates:
$x = -1 + 2 = 1$.
$y = -(-1) - 3 = 1 - 3 = -2$.
$z = -6(-1) + 1 = 6 + 1 = 7$.
Thus,the point of intersection is $(1, -2, 7)$.

(C) The three planes pass through a common line. This means the family of planes passing through the intersection of any two planes must contain the third plane.
Consider the planes $P_1: x - 5 = 0$,$P_2: 2x - 5ay + 3z - 2 = 0$,and $P_3: 3bx + y - 3z = 0$.
The equation of any plane passing through the intersection of $P_1$ and $P_2$ is given by $P_1 + \lambda P_2 = 0$.
$(x - 5) + \lambda(2x - 5ay + 3z - 2) = 0$
$(1 + 2\lambda)x - (5a\lambda)y + (3\lambda)z - (5 + 2\lambda) = 0$.
Since this plane is the same as $P_3$,we compare the coefficients with $3bx + y - 3z = 0$.
For the coefficients to be proportional:
$\frac{1 + 2\lambda}{3b} = \frac{-5a\lambda}{1} = \frac{3\lambda}{-3} = \frac{-(5 + 2\lambda)}{0}$.
From $\frac{3\lambda}{-3} = -\lambda$,we have the ratio $k = -\lambda$.
From $\frac{-(5 + 2\lambda)}{0}$,the denominator must be zero,so $0 = 0$ is not possible unless the constant term is also zero. However,looking at the ratio $\frac{3\lambda}{-3} = -\lambda$,we equate:
$-5a\lambda = 1 \implies a = -\frac{1}{5\lambda}$.
$3\lambda = -3 \implies \lambda = -1$.
Substituting $\lambda = -1$ into the ratio:
$a = -\frac{1}{5(-1)} = \frac{1}{5}$.
Now for $b$:
$\frac{1 + 2(-1)}{3b} = -(-1) = 1
\implies \frac{-1}{3b} = 1 \implies b = -\frac{1}{3}$.
Wait,let us re-evaluate the constant term. The plane $P_3$ has no constant term,so the plane $P_1 + \lambda P_2$ must have a constant term of $0$.
$-(5 + 2\lambda) = 0 \implies \lambda = -\frac{5}{2}$.
Now substitute $\lambda = -\frac{5}{2}$ into the other ratios:
$\frac{1 + 2(-5/2)}{3b} = \frac{-5a(-5/2)}{1} = \frac{3(-5/2)}{-3} = 1$.
$\frac{1 - 5}{3b} = 1 \implies \frac{-4}{3b} = 1 \implies b = -\frac{4}{3}$.
$\frac{25a}{2} = 1 \implies a = \frac{2}{25}$.
Re-checking the question,if $x=5$ is a plane,it is $x-5=0$. The intersection of $x=5$ and $2x-5ay+3z-2=0$ gives $2(5)-5ay+3z-2=0 \implies -5ay+3z+8=0$.
Comparing with $3bx+y-3z=0$ at $x=5$: $15b+y-3z=0$.
Since they share a line,$y-3z = -15b$ and $-5ay+3z = -8$.
Adding these: $(1-5a)y = -8-15b$. For this to hold for all points on the line,coefficients must match: $a = 1/5$ and $b = -8/15$.
Thus,$(a, b) = (1/5, -8/15)$.

(B) Let $\frac{x}{1} = \frac{y}{2} = \frac{z}{2} = k$.
Then,we have $x = k$,$y = 2k$,and $z = 2k$.
Any point on the line is of the form $(k, 2k, 2k)$.
Since this point lies on the plane $2x + y + z = 6$,we substitute the coordinates into the equation:
$2(k) + (2k) + (2k) = 6$.
$6k = 6$.
$k = 1$.
Substituting $k = 1$ back into the coordinates $(k, 2k, 2k)$,we get the point of intersection as $(1, 2, 2)$.

(A) Let the plane $ax + by + cz + d = 0$ divide the line segment joining the points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ in the ratio $k:1$.
Using the section formula,the coordinates of the point of division $P$ are:
$P = \left( \frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1}, \frac{kz_2 + z_1}{k+1} \right)$.
Since this point $P$ lies on the plane $ax + by + cz + d = 0$,it must satisfy the equation:
$a\left( \frac{kx_2 + x_1}{k+1} \right) + b\left( \frac{ky_2 + y_1}{k+1} \right) + c\left( \frac{kz_2 + z_1}{k+1} \right) + d = 0$.
Multiplying by $(k+1)$,we get:
$a(kx_2 + x_1) + b(ky_2 + y_1) + c(kz_2 + z_1) + d(k+1) = 0$.
Rearranging the terms to solve for $k$:
$k(ax_2 + by_2 + cz_2 + d) + (ax_1 + by_1 + cz_1 + d) = 0$.
$k(ax_2 + by_2 + cz_2 + d) = -(ax_1 + by_1 + cz_1 + d)$.
$k = -\frac{ax_1 + by_1 + cz_1 + d}{ax_2 + by_2 + cz_2 + d}$.
Thus,the plane divides the line segment in the ratio $-\frac{ax_1 + by_1 + cz_1 + d}{ax_2 + by_2 + cz_2 + d} : 1$.

(D) Let the line be $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3} = r$.
Then,the coordinates of any point on the line are given by $x = r$,$y = 2r + 1$,and $z = 3r - 2$.
Since this point lies on the plane $2x + 3y + z = 0$,we substitute these values into the plane equation:
$2(r) + 3(2r + 1) + (3r - 2) = 0$.
Expanding the equation:
$2r + 6r + 3 + 3r - 2 = 0$.
Combining like terms:
$11r + 1 = 0$,which gives $r = -\frac{1}{11}$.
Now,substitute $r = -\frac{1}{11}$ back into the expressions for $x, y, z$:
$x = -\frac{1}{11}$,
$y = 2(-\frac{1}{11}) + 1 = -\frac{2}{11} + \frac{11}{11} = \frac{9}{11}$,
$z = 3(-\frac{1}{11}) - 2 = -\frac{3}{11} - \frac{22}{11} = -\frac{25}{11}$.
Thus,the point of intersection is $\left( -\frac{1}{11}, \frac{9}{11}, -\frac{25}{11} \right)$.

(D) Let the image of the point $P(-1, 3, 4)$ in the plane $x - 2y = 0$ be $P'(\alpha, \beta, \gamma)$.
The line passing through $P$ and $P'$ is perpendicular to the plane $x - 2y = 0$. The normal vector to the plane is $\vec{n} = (1, -2, 0)$.
The equation of the line passing through $P(-1, 3, 4)$ with direction ratios $(1, -2, 0)$ is $\frac{x + 1}{1} = \frac{y - 3}{-2} = \frac{z - 4}{0} = k$.
Thus,any point on this line is $(k - 1, -2k + 3, 4)$.
The midpoint of $PP'$ is $M = (\frac{\alpha - 1}{2}, \frac{\beta + 3}{2}, \frac{\gamma + 4}{2})$.
Since $M$ lies on the plane $x - 2y = 0$,we have $\frac{\alpha - 1}{2} - 2(\frac{\beta + 3}{2}) = 0$,which simplifies to $\alpha - 1 - 2\beta - 6 = 0$,or $\alpha - 2\beta = 7$.
Also,the line $PP'$ is perpendicular to the plane,so the direction ratios of $PP'$ are proportional to the normal of the plane: $\frac{\alpha - (-1)}{1} = \frac{\beta - 3}{-2} = \frac{\gamma - 4}{0} = \lambda$.
From $\frac{\gamma - 4}{0} = \lambda$,we get $\gamma = 4$.
From $\alpha + 1 = \lambda$ and $\beta - 3 = -2\lambda$,we have $\alpha = \lambda - 1$ and $\beta = -2\lambda + 3$.
Substituting into $\alpha - 2\beta = 7$: $(\lambda - 1) - 2(-2\lambda + 3) = 7 \Rightarrow \lambda - 1 + 4\lambda - 6 = 7 \Rightarrow 5\lambda = 14 \Rightarrow \lambda = \frac{14}{5}$.
Then $\alpha = \frac{14}{5} - 1 = \frac{9}{5}$ and $\beta = -2(\frac{14}{5}) + 3 = -\frac{28}{5} + \frac{15}{5} = -\frac{13}{5}$.
The image is $(\frac{9}{5}, -\frac{13}{5}, 4)$. Since this is not among the options,the correct choice is $D$.

(B) Let the direction ratios of the line $L$ be $(a, b, c)$.
Since the line $L$ lies in both planes,it is perpendicular to the normals of both planes.
The normal vectors are $\vec{n_1} = (2, 3, 1)$ and $\vec{n_2} = (1, 3, 2)$.
The direction vector $\vec{v}$ of the line $L$ is given by the cross product $\vec{n_1} \times \vec{n_2}$:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{vmatrix} = \hat{i}(6-3) - \hat{j}(4-1) + \hat{k}(6-3) = 3\hat{i} - 3\hat{j} + 3\hat{k}$.
We can simplify the direction vector to $(1, -1, 1)$.
The direction cosines $(\ell, m, n)$ are obtained by normalizing the vector $(1, -1, 1)$:
Magnitude $= \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
Thus,$\ell = \frac{1}{\sqrt{3}}$,$m = -\frac{1}{\sqrt{3}}$,$n = \frac{1}{\sqrt{3}}$.
The angle $\alpha$ with the positive $x$-axis satisfies $\cos \alpha = \ell = \frac{1}{\sqrt{3}}$.

(A) The line $\frac{x - 2}{3} = \frac{y - 1}{-5} = \frac{z + 2}{2}$ lies on the plane $x + 3y - \alpha z + \beta = 0$.
Since the line is parallel to the plane,the normal vector of the plane $(1, 3, -\alpha)$ must be perpendicular to the direction vector of the line $(3, -5, 2)$.
Thus,$1(3) + 3(-5) + (-\alpha)(2) = 0$.
$3 - 15 - 2\alpha = 0$.
$-12 - 2\alpha = 0 \implies \alpha = -6$.
Now,the plane equation is $x + 3y + 6z + \beta = 0$.
Since the line lies on the plane,any point on the line must satisfy the plane equation. The point $(2, 1, -2)$ lies on the line.
Substituting $(2, 1, -2)$ into the plane equation: $2 + 3(1) + 6(-2) + \beta = 0$.
$2 + 3 - 12 + \beta = 0$.
$-7 + \beta = 0 \implies \beta = 7$.
Therefore,$(\alpha, \beta) = (-6, 7)$.

(D) The line is given by $\frac{x-0}{1} = \frac{y-1}{2} = \frac{z-3}{\lambda}$. The direction vector of the line is $\vec{b} = (1, 2, \lambda)$.
The normal vector to the plane $x + 2y + 3z = 4$ is $\vec{n} = (1, 2, 3)$.
Let $\theta$ be the angle between the line and the plane. The formula for the angle is $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Given $\theta = \cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,we have $\cos \theta = \sqrt{\frac{5}{14}}$.
Using $\sin^2 \theta + \cos^2 \theta = 1$,we get $\sin \theta = \sqrt{1 - \frac{5}{14}} = \sqrt{\frac{9}{14}} = \frac{3}{\sqrt{14}}$.
Now,$\frac{|(1)(1) + (2)(2) + (3)(\lambda)|}{\sqrt{1^2 + 2^2 + \lambda^2} \sqrt{1^2 + 2^2 + 3^2}} = \frac{3}{\sqrt{14}}$.
$\frac{|1 + 4 + 3\lambda|}{\sqrt{5 + \lambda^2} \sqrt{14}} = \frac{3}{\sqrt{14}}$.
$|5 + 3\lambda| = 3\sqrt{5 + \lambda^2}$.
Squaring both sides: $(5 + 3\lambda)^2 = 9(5 + \lambda^2)$.
$25 + 9\lambda^2 + 30\lambda = 45 + 9\lambda^2$.
$30\lambda = 20$.
$\lambda = \frac{20}{30} = \frac{2}{3}$.

(C) Let the two lines be represented by parameters $\lambda$ and $\mu$ respectively:
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda \implies x=2\lambda+1, y=3\lambda-1, z=4\lambda+1$
$\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu \implies x=\mu+3, y=2\mu+k, z=\mu$
If the lines intersect,there must exist a point common to both,so:
$2\lambda+1 = \mu+3 \implies 2\lambda - \mu = 2 \quad (i)$
$3\lambda-1 = 2\mu+k \implies 3\lambda - 2\mu = k+1 \quad (ii)$
$4\lambda+1 = \mu \implies 4\lambda - \mu = -1 \quad (iii)$
Subtracting $(i)$ from $(iii)$:
$(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2
\implies 2\lambda = -3 \implies \lambda = -\frac{3}{2}$
Substituting $\lambda = -\frac{3}{2}$ into $(iii)$:
$4(-\frac{3}{2}) + 1 = \mu \implies -6 + 1 = \mu \implies \mu = -5$
Substituting $\lambda = -\frac{3}{2}$ and $\mu = -5$ into $(ii)$:
$3(-\frac{3}{2}) - 2(-5) = k+1
\implies -\frac{9}{2} + 10 = k+1
\implies k = 9 - \frac{9}{2} = \frac{9}{2}$

(C) Two lines $\frac{x-x_1}{l_1} = \frac{y-y_1}{m_1} = \frac{z-z_1}{n_1}$ and $\frac{x-x_2}{l_2} = \frac{y-y_2}{m_2} = \frac{z-z_2}{n_2}$ are coplanar if and only if the scalar triple product of the vector connecting the points and the direction vectors is zero,i.e.,$[(x_2-x_1, y_2-y_1, z_2-z_1), (l_1, m_1, n_1), (l_2, m_2, n_2)] = 0$.
Given points are $P_1(2, 3, 4)$ and $P_2(1, 4, 5)$. The vector $\vec{P_1P_2} = (1-2, 4-3, 5-4) = (-1, 1, 1)$.
The direction vectors are $\vec{v_1} = (1, 1, -k)$ and $\vec{v_2} = (k, 2, 1)$.
The condition for coplanarity is:
$\left| \begin{matrix} 1-2 & 4-3 & 5-4 \\ 1 & 1 & -k \\ k & 2 & 1 \end{matrix} \right| = 0$
$\left| \begin{matrix} -1 & 1 & 1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{matrix} \right| = 0$
Expanding the determinant:
$-1(1 + 2k) - 1(1 + k^2) + 1(2 - k) = 0$
$-1 - 2k - 1 - k^2 + 2 - k = 0$
$-k^2 - 3k = 0$
$k^2 + 3k = 0$
$k(k + 3) = 0$
Thus,$k = 0$ or $k = -3$. There are exactly two values for $k$.

(C) Let the given line be $L_1: \frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{-5} = k$. Any point on $L_1$ is $P(3k + 1, k + 3, -5k + 4)$.
To find the intersection point $B$ of $L_1$ and the plane $2x - y + z + 3 = 0$,substitute the coordinates of $P$ into the plane equation:
$2(3k + 1) - (k + 3) + (-5k + 4) + 3 = 0$
$6k + 2 - k - 3 - 5k + 4 + 3 = 0$
$6 = 0$,which is impossible. This implies the line is parallel to the plane.
Let $A(1, 3, 4)$ be a point on the line. The image $A'$ of $A$ in the plane is given by $\frac{x - 1}{2} = \frac{y - 3}{-1} = \frac{z - 4}{1} = -2 \frac{2(1) - 3 + 4 + 3}{2^2 + (-1)^2 + 1^2} = -2 \frac{6}{6} = -2$.
So,$x - 1 = -4 \Rightarrow x = -3$,$y - 3 = 2 \Rightarrow y = 5$,$z - 4 = -2 \Rightarrow z = 2$. Thus,$A'(-3, 5, 2)$.
The image line passes through $A'(-3, 5, 2)$ and is parallel to the original line,so its direction ratios are $(3, 1, -5)$.
The equation of the image line is $\frac{x + 3}{3} = \frac{y - 5}{1} = \frac{z - 2}{-5}$.

(A) Let the line be $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} = \lambda$.
Then,any point on the line is given by $(3\lambda + 2, 4\lambda - 1, 12\lambda + 2)$.
Since this point lies on the plane $x - y + z = 16$,we substitute these coordinates into the plane equation:
$(3\lambda + 2) - (4\lambda - 1) + (12\lambda + 2) = 16$
$3\lambda + 2 - 4\lambda + 1 + 12\lambda + 2 = 16$
$11\lambda + 5 = 16$
$11\lambda = 11$
$\lambda = 1$
Substituting $\lambda = 1$ back into the coordinates,we get the point of intersection:
$x = 3(1) + 2 = 5$
$y = 4(1) - 1 = 3$
$z = 12(1) + 2 = 14$
So,the point of intersection is $(5, 3, 14)$.
Now,we find the distance between the point $(1, 0, 2)$ and $(5, 3, 14)$ using the distance formula:
$d = \sqrt{(5 - 1)^2 + (3 - 0)^2 + (14 - 2)^2}$
$d = \sqrt{4^2 + 3^2 + 12^2}$
$d = \sqrt{16 + 9 + 144}$
$d = \sqrt{169} = 13$.
Thus,the distance is $13$ units.

(D) The equation of the family of planes passing through the intersection of the planes $2x - 5y + z - 3 = 0$ and $x + y + 4z - 5 = 0$ is given by:
$(2x - 5y + z - 3) + \lambda(x + y + 4z - 5) = 0$
Rearranging the terms,we get:
$(2 + \lambda)x + (\lambda - 5)y + (4\lambda + 1)z - (3 + 5\lambda) = 0 \quad \dots(i)$
Since this plane is parallel to the plane $x + 3y + 6z = 1$,the normal vectors must be proportional:
$\frac{2 + \lambda}{1} = \frac{\lambda - 5}{3} = \frac{4\lambda + 1}{6} = k$
From $\frac{2 + \lambda}{1} = \frac{\lambda - 5}{3}$,we have $6 + 3\lambda = \lambda - 5 \implies 2\lambda = -11 \implies \lambda = -\frac{11}{2}$.
Substituting $\lambda = -\frac{11}{2}$ into equation $(i)$:
$(2 - \frac{11}{2})x + (-\frac{11}{2} - 5)y + (4(-\frac{11}{2}) + 1)z - (3 + 5(-\frac{11}{2})) = 0$
$-\frac{7}{2}x - \frac{21}{2}y - 21z - (3 - \frac{55}{2}) = 0$
$-\frac{7}{2}x - \frac{21}{2}y - 21z + \frac{49}{2} = 0$
Multiplying by $-\frac{2}{7}$,we get:
$x + 3y + 6z - 7 = 0 \implies x + 3y + 6z = 7$.

(B) The given line is $\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z+4}{3}$.
The direction vector of the line is $\vec{v} = (2, -1, 3)$ and a point on the line is $P(3, -2, -4)$.
The equation of the plane is $lx + my - z = 9$,which has a normal vector $\vec{n} = (l, m, -1)$.
Since the line lies in the plane,the normal vector $\vec{n}$ must be perpendicular to the direction vector $\vec{v}$ of the line. Thus,$\vec{n} \cdot \vec{v} = 0$:
$2l - m - 3 = 0 \Rightarrow 2l - m = 3$ ....$(1)$
Also,the point $P(3, -2, -4)$ must satisfy the equation of the plane:
$l(3) + m(-2) - (-4) = 9$
$3l - 2m + 4 = 9$
$3l - 2m = 5$ ....$(2)$
Solving equations $(1)$ and $(2)$:
From $(1)$,$m = 2l - 3$. Substituting into $(2)$:
$3l - 2(2l - 3) = 5$
$3l - 4l + 6 = 5$
$-l = -1 \Rightarrow l = 1$
Substituting $l = 1$ into $(1)$:
$m = 2(1) - 3 = -1$
Therefore,$l^2 + m^2 = (1)^2 + (-1)^2 = 1 + 1 = 2$.

(D) The equation of the line passing through $(1, -5, 9)$ and parallel to the line $x = y = z$ is given by $\frac{x - 1}{1} = \frac{y + 5}{1} = \frac{z - 9}{1} = \lambda$.
Any point on this line is of the form $P(\lambda + 1, \lambda - 5, \lambda + 9)$.
Since this point $P$ lies on the plane $x - y + z = 5$,we substitute the coordinates into the plane equation:
$(\lambda + 1) - (\lambda - 5) + (\lambda + 9) = 5$
$\lambda + 1 - \lambda + 5 + \lambda + 9 = 5$
$\lambda + 15 = 5$
$\lambda = -10$.
Substituting $\lambda = -10$ back into the coordinates of $P$,we get:
$P = (-10 + 1, -10 - 5, -10 + 9) = (-9, -15, -1)$.
The distance between the points $(1, -5, 9)$ and $(-9, -15, -1)$ is calculated using the distance formula:
$d = \sqrt{(-9 - 1)^2 + (-15 - (-5))^2 + (-1 - 9)^2}$
$d = \sqrt{(-10)^2 + (-10)^2 + (-10)^2}$
$d = \sqrt{100 + 100 + 100} = \sqrt{300} = 10\sqrt{3}$.

(C) The equation of the line passing through $P(1, -2, 3)$ and parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is given by $\frac{x-1}{1} = \frac{y+2}{4} = \frac{z-3}{5} = \lambda$.
Any point $F$ on this line can be represented as $(\lambda+1, 4\lambda-2, 5\lambda+3)$.
Since $F$ lies on the plane $2x + 3y - 4z + 22 = 0$,we substitute the coordinates of $F$ into the plane equation:
$2(\lambda+1) + 3(4\lambda-2) - 4(5\lambda+3) + 22 = 0$
$2\lambda + 2 + 12\lambda - 6 - 20\lambda - 12 + 22 = 0$
$-6\lambda + 6 = 0 \Rightarrow \lambda = 1$.
Substituting $\lambda = 1$ back into the coordinates of $F$,we get $F(2, 2, 8)$.
Since $F$ is the midpoint of $PQ$,the distance $PQ = 2PF$.
The distance $PF = \sqrt{(2-1)^2 + (2-(-2))^2 + (8-3)^2} = \sqrt{1^2 + 4^2 + 5^2} = \sqrt{1 + 16 + 25} = \sqrt{42}$.
Therefore,$PQ = 2PF = 2\sqrt{42}$.

(C) Let the equation of the plane passing through $(1, -1, -1)$ be $a(x - 1) + b(y + 1) + c(z + 1) = 0$.
The normal vector $\vec{n}$ is perpendicular to the direction vectors of the two lines,$\vec{v_1} = (1, -2, 3)$ and $\vec{v_2} = (2, -1, -1)$.
$\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{vmatrix} = \hat{i}(2 + 3) - \hat{j}(-1 - 6) + \hat{k}(-1 + 4) = 5\hat{i} + 7\hat{j} + 3\hat{k}$.
Thus,the equation of the plane is $5(x - 1) + 7(y + 1) + 3(z + 1) = 0$,which simplifies to $5x + 7y + 3z + 5 = 0$.
The distance of the point $(1, 3, -7)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
Substituting the values,$d = \frac{|5(1) + 7(3) + 3(-7) + 5|}{\sqrt{5^2 + 7^2 + 3^2}} = \frac{|5 + 21 - 21 + 5|}{\sqrt{25 + 49 + 9}} = \frac{10}{\sqrt{83}}$.

(A) The equation of the plane passing through the line of intersection of the planes $2x - 2y + 3z - 2 = 0$ and $x - y + z + 1 = 0$ is given by:
$(2x - 2y + 3z - 2) + \lambda(x - y + z + 1) = 0$
$x(\lambda + 2) - y(\lambda + 2) + z(\lambda + 3) + (\lambda - 2) = 0 \quad \dots(1)$
Since this plane contains the line ${L_2}$ formed by the intersection of $x + 2y - z - 3 = 0$ and $3x - y + 2z - 1 = 0$,the normal vector of the plane must be perpendicular to the direction vectors of the lines forming ${L_2}$. Alternatively,the plane must satisfy the condition that the determinant of the coefficients of the three planes is zero:
$\begin{vmatrix} \lambda + 2 & -(\lambda + 2) & \lambda + 3 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} = 0$
Expanding the determinant:
$(\lambda + 2)(4 - 1) + (\lambda + 2)(2 + 3) + (\lambda + 3)(-1 - 6) = 0$
$3(\lambda + 2) + 5(\lambda + 2) - 7(\lambda + 3) = 0$
$8\lambda + 16 - 7\lambda - 21 = 0 \Rightarrow \lambda = 5$
Substituting $\lambda = 5$ into equation $(1)$:
$x(5 + 2) - y(5 + 2) + z(5 + 3) + (5 - 2) = 0$
$7x - 7y + 8z + 3 = 0$
The perpendicular distance from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.
$d = \frac{|3|}{\sqrt{7^2 + (-7)^2 + 8^2}} = \frac{3}{\sqrt{49 + 49 + 64}} = \frac{3}{\sqrt{162}} = \frac{3}{9\sqrt{2}} = \frac{1}{3\sqrt{2}}$.

(C) Let the points be $A(4, -1, 3)$ and $B(5, -1, 4)$. The vector $\overrightarrow{AB} = (5-4)\hat{i} + (-1 - (-1))\hat{j} + (4-3)\hat{k} = \hat{i} + \hat{k}$.
The normal to the plane $x + y + z = 7$ is $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.
The unit normal vector is $\hat{n} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.
The projection of $\overrightarrow{AB}$ on the normal $\vec{n}$ is $d = |\overrightarrow{AB} \cdot \hat{n}| = |(\hat{i} + \hat{k}) \cdot \frac{(\hat{i} + \hat{j} + \hat{k})}{\sqrt{3}}| = \frac{1+0+1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$.
The length of the projection of the line segment $AB$ on the plane is given by $\sqrt{|\overrightarrow{AB}|^2 - d^2}$.
Here,$|\overrightarrow{AB}|^2 = 1^2 + 0^2 + 1^2 = 2$.
So,the length of the projection $= \sqrt{2 - (\frac{2}{\sqrt{3}})^2} = \sqrt{2 - \frac{4}{3}} = \sqrt{\frac{6-4}{3}} = \sqrt{\frac{2}{3}}$.

(C) The angle $\theta$ between a line with direction vector $\vec{b} = (1, 2, 2)$ and a plane with normal vector $\vec{n} = (2, -1, \sqrt{\lambda})$ is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Given $\sin \theta = \frac{1}{3}$,we have $\frac{1}{3} = \frac{|(1)(2) + (2)(-1) + (2)(\sqrt{\lambda})|}{\sqrt{1^2 + 2^2 + 2^2} \sqrt{2^2 + (-1)^2 + (\sqrt{\lambda})^2}}$.
Simplifying the expression: $\frac{1}{3} = \frac{|2 - 2 + 2\sqrt{\lambda}|}{\sqrt{9} \sqrt{5 + \lambda}}$.
$\frac{1}{3} = \frac{2\sqrt{\lambda}}{3\sqrt{5 + \lambda}}$.
Squaring both sides: $\frac{1}{9} = \frac{4\lambda}{9(5 + \lambda)}$.
$5 + \lambda = 4\lambda$.
$3\lambda = 5 \Rightarrow \lambda = \frac{5}{3}$.