Line and Plane Questions in English

(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 $\frac{x-1}{-3} = \frac{y-2}{5} = \frac{z-3}{4}$.

(C) Let the direction ratios of the required line be $a, b, c$. Since the line lies in both planes,it is perpendicular to the normals of both planes. Thus,$3a + 2b + c = 0$ and $a + b - 2c = 0$.
Using the cross product method,we have:
$\frac{a}{(2)(-2) - (1)(1)} = \frac{b}{(1)(1) - (3)(-2)} = \frac{c}{(3)(1) - (2)(1)}$
$\frac{a}{-4-1} = \frac{b}{1+6} = \frac{c}{3-2}$
$\frac{a}{-5} = \frac{b}{7} = \frac{c}{1}$.
To find a point on the line,we set $z = 0$ in the given plane equations:
$3x + 2y = 5$ and $x + y = 3$.
Solving these,we multiply the second by $2$: $2x + 2y = 6$.
Subtracting this from the first: $(3x - 2x) = 5 - 6 \Rightarrow x = -1$.
Substituting $x = -1$ into $x + y = 3$,we get $-1 + y = 3 \Rightarrow y = 4$.
So,the point is $(-1, 4, 0)$.
The symmetric equation is $\frac{x - (-1)}{-5} = \frac{y - 4}{7} = \frac{z - 0}{1}$,which is $\frac{x+1}{-5} = \frac{y-4}{7} = \frac{z-0}{1}$.

(A) The equation of any plane passing through the line of intersection of the planes $P_1: x+y+z-1=0$ and $P_2: 3x+4y+5z-2=0$ is given by $P_1 + \lambda P_2 = 0$.
$(x+y+z-1) + \lambda(3x+4y+5z-2) = 0$
$(1+3\lambda)x + (1+4\lambda)y + (1+5\lambda)z - (1+2\lambda) = 0$.
This plane is perpendicular to the $XY$-plane. The equation of the $XY$-plane is $z=0$,and its normal vector is $\vec{n_1} = (0, 0, 1)$.
The normal vector of our required plane is $\vec{n_2} = (1+3\lambda, 1+4\lambda, 1+5\lambda)$.
Since the planes are perpendicular,the dot product of their normal vectors must be zero:
$\vec{n_1} \cdot \vec{n_2} = 0 \implies (0)(1+3\lambda) + (0)(1+4\lambda) + (1)(1+5\lambda) = 0$.
$1+5\lambda = 0 \implies \lambda = -\frac{1}{5}$.
Substituting $\lambda = -\frac{1}{5}$ into the equation:
$(x+y+z-1) - \frac{1}{5}(3x+4y+5z-2) = 0$.
$5(x+y+z-1) - (3x+4y+5z-2) = 0$.
$5x+5y+5z-5 - 3x-4y-5z+2 = 0$.
$2x+y-3=0$.

(B) The equation of the family of planes passing through the intersection of $P_1: 2x-y+z-3=0$ and $P_2: 4x-3y+5z+9=0$ is given by $P_1 + \lambda P_2 = 0$.
$(2x-y+z-3) + \lambda(4x-3y+5z+9) = 0$
$(2+4\lambda)x + (-1-3\lambda)y + (1+5\lambda)z + (-3+9\lambda) = 0$.
Since the plane is parallel to the line with direction ratios $(2, 4, 5)$,the normal vector of the plane is perpendicular to the line's direction vector.
Therefore,$2(2+4\lambda) + 4(-1-3\lambda) + 5(1+5\lambda) = 0$.
$4 + 8\lambda - 4 - 12\lambda + 5 + 25\lambda = 0$.
$21\lambda + 5 = 0 \implies \lambda = -\frac{5}{21}$.
Substituting $\lambda$ into the equation:
$(2 - \frac{20}{21})x + (-1 + \frac{15}{21})y + (1 - \frac{25}{21})z + (-3 - \frac{45}{21}) = 0$.
$(\frac{42-20}{21})x + (\frac{-21+15}{21})y + (\frac{21-25}{21})z + (\frac{-63-45}{21}) = 0$.
$22x - 6y - 4z - 108 = 0$.
Dividing by $2$: $11x - 3y - 2z - 54 = 0$.
Here $\alpha=11, \beta=-3, \gamma=-2, d=-54$.
$\alpha+\beta+\gamma+d = 11 - 3 - 2 - 54 = -48$.

(D) The distance of point $P(x_1, y_1, z_1) = (1, -2, 1)$ from the plane $Ax + By + Cz - D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}}$.
Given the plane $x + 2y - 2z - \alpha = 0$,we have $A=1, B=2, C=-2, D=\alpha$.
Substituting the values: $5 = \frac{|1(1) + 2(-2) - 2(1) - \alpha|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{|1 - 4 - 2 - \alpha|}{\sqrt{9}} = \frac{|-5 - \alpha|}{3}$.
Since $\alpha > 0$,$|-5 - \alpha| = 5 + \alpha$. Thus,$5 = \frac{5 + \alpha}{3} \implies 15 = 5 + \alpha \implies \alpha = 10$.
The plane equation is $x + 2y - 2z = 10$.
The line passing through $P(1, -2, 1)$ perpendicular to the plane has direction ratios $(1, 2, -2)$. The equation of the line is $\frac{x-1}{1} = \frac{y+2}{2} = \frac{z-1}{-2} = k$.
Any point on this line is $(k+1, 2k-2, -2k+1)$.
Since this point lies on the plane: $(k+1) + 2(2k-2) - 2(-2k+1) = 10$.
$k + 1 + 4k - 4 + 4k - 2 = 10 \implies 9k - 5 = 10 \implies 9k = 15 \implies k = \frac{5}{3}$.
Substituting $k = \frac{5}{3}$ into the point coordinates: $x = \frac{5}{3} + 1 = \frac{8}{3}$,$y = 2(\frac{5}{3}) - 2 = \frac{4}{3}$,$z = -2(\frac{5}{3}) + 1 = -\frac{7}{3}$.
The foot of the perpendicular is $\left(\frac{8}{3}, \frac{4}{3}, -\frac{7}{3}\right)$.

(A) Let the line be $L: \frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3} = k$. Any point $Q$ on the line is $(2k+1, 4k+3, 3k+2)$.
The vector $\vec{PQ} = (2k+1-3, 4k+3-8, 3k+2-2) = (2k-2, 4k-5, 3k)$.
The line is measured parallel to the plane $3x + 2y - 2z + 15 = 0$. The normal to the plane is $\vec{n} = (3, 2, -2)$.
Since the line segment $PQ$ is parallel to the plane, it must be perpendicular to the normal vector $\vec{n}$.
Thus, $\vec{PQ} \cdot \vec{n} = 0$.
$(2k-2)(3) + (4k-5)(2) + (3k)(-2) = 0$.
$6k - 6 + 8k - 10 - 6k = 0$.
$8k - 16 = 0 \implies k = 2$.
Substituting $k=2$ into $\vec{PQ}$, we get $\vec{PQ} = (2(2)-2, 4(2)-5, 3(2)) = (2, 3, 6)$.
The distance is the magnitude of $\vec{PQ} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \text{ units}$.

(A) The equation of the line is $\bar{r} = \bar{a} + \lambda \bar{b}$,where $\bar{a} = 3\hat{i} - 2\hat{j} + \hat{k}$ and $\bar{b} = \hat{i} - 2\hat{j}$.
The equation of the plane is $\bar{r} \cdot \bar{n} = d$,where $\bar{n} = 2\hat{i} + \hat{j} + \hat{k}$ and $d = 4$.
First,check if the line is parallel to the plane by calculating $\bar{b} \cdot \bar{n} = (1)(2) + (-2)(1) + (0)(1) = 2 - 2 + 0 = 0$.
Since $\bar{b} \cdot \bar{n} = 0$,the line is parallel to the plane.
The distance $D$ between a parallel line and a plane is given by $D = \frac{|\bar{a} \cdot \bar{n} - d|}{|\bar{n}|}$.
Calculate $\bar{a} \cdot \bar{n} = (3)(2) + (-2)(1) + (1)(1) = 6 - 2 + 1 = 5$.
Calculate $|\bar{n}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
Thus,$D = \frac{|5 - 4|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$ units.

(C) The equation of a plane passing through the intersection of two planes $P_1: x+2y-z+1=0$ and $P_2: 3x-y-4z+3=0$ is given by $P_1 + \lambda P_2 = 0$.
$(x+2y-z+1) + \lambda(3x-y-4z+3) = 0$.
Since the plane passes through the point $(1,1,1)$,we substitute $x=1, y=1, z=1$ into the equation:
$(1+2(1)-1+1) + \lambda(3(1)-1-4(1)+3) = 0$.
$(1+2-1+1) + \lambda(3-1-4+3) = 0$.
$3 + \lambda(1) = 0 \implies \lambda = -3$.
Substituting $\lambda = -3$ back into the equation:
$(x+2y-z+1) - 3(3x-y-4z+3) = 0$.
$x+2y-z+1 - 9x+3y+12z-9 = 0$.
$-8x + 5y + 11z - 8 = 0$.
Multiplying by $-1$,we get $8x - 5y - 11z + 8 = 0$.

(C) The equation of a plane passing through the intersection of two planes $P_1: \overline{r} \cdot \overline{n}_1 = d_1$ and $P_2: \overline{r} \cdot \overline{n}_2 = d_2$ is given by $(\overline{r} \cdot \overline{n}_1 - d_1) + \lambda(\overline{r} \cdot \overline{n}_2 - d_2) = 0$.
Given planes are $P_1: \overline{r} \cdot(2 \hat{i}-3 \hat{j}+4 \hat{k}) - 1 = 0$ and $P_2: \overline{r} \cdot(\hat{i}-\hat{j}) + 4 = 0$.
The equation of the required plane is $\overline{r} \cdot((2+\lambda) \hat{i} + (-3-\lambda) \hat{j} + 4 \hat{k}) - 1 + 4\lambda = 0$.
This plane is perpendicular to $\overline{r} \cdot(2 \hat{i}-\hat{j}+\hat{k}) + 8 = 0$.
Therefore,the dot product of their normal vectors is zero: $(2+\lambda)(2) + (-3-\lambda)(-1) + (4)(1) = 0$.
$4 + 2\lambda + 3 + \lambda + 4 = 0 \implies 3\lambda + 11 = 0 \implies \lambda = -\frac{11}{3}$.
Substituting $\lambda$ into the plane equation: $\overline{r} \cdot((2-\frac{11}{3}) \hat{i} + (-3+\frac{11}{3}) \hat{j} + 4 \hat{k}) - 1 + 4(-\frac{11}{3}) = 0$.
$\overline{r} \cdot(-\frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + 4 \hat{k}) - 1 - \frac{44}{3} = 0$.
Multiply by $3$: $\overline{r} \cdot(-5 \hat{i} + 2 \hat{j} + 12 \hat{k}) - 3 - 44 = 0$.
$\overline{r} \cdot(-5 \hat{i} + 2 \hat{j} + 12 \hat{k}) = 47$.
Comparing with $\overline{r} \cdot(-5 \hat{i} + 2 \hat{j} + 12 \hat{k}) = \mu$,we get $\mu = 47$.

(C) The normal vector to the plane $x-y+z=3$ is $\vec{n} = (1, -1, 1)$.
Let $Q = (3, 1, 7)$. The line passing through $Q$ and perpendicular to the plane is given by $\frac{x-3}{1} = \frac{y-1}{-1} = \frac{z-7}{1} = \lambda$.
Any point on this line is $(\lambda+3, -\lambda+1, \lambda+7)$.
For the intersection point $M$ with the plane,we have $(\lambda+3) - (-\lambda+1) + (\lambda+7) = 3$,which simplifies to $3\lambda + 9 = 3$,so $3\lambda = -6$,giving $\lambda = -2$.
Thus,$M = (-2+3, -(-2)+1, -2+7) = (1, 3, 5)$.
Since $M$ is the midpoint of $PQ$,if $P = (a, b, c)$,then $\frac{3+a}{2} = 1, \frac{1+b}{2} = 3, \frac{7+c}{2} = 5$.
This gives $a = -1, b = 5, c = 3$,so $P = (-1, 5, 3)$.
The plane passes through $P(-1, 5, 3)$ and contains the line $\frac{x}{1} = \frac{y}{2} = \frac{z}{1}$,which passes through the origin $O(0, 0, 0)$ with direction vector $\vec{v} = (1, 2, 1)$.
The normal to the required plane is $\vec{n'} = \vec{OP} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 5 & 3 \\ 1 & 2 & 1 \end{vmatrix} = \hat{i}(5-6) - \hat{j}(-1-3) + \hat{k}(-2-5) = -\hat{i} + 4\hat{j} - 7\hat{k}$.
The equation of the plane is $-1(x-0) + 4(y-0) - 7(z-0) = 0$,which is $-x + 4y - 7z = 0$,or $x - 4y + 7z = 0$.

(C) Let the required plane be $P_1$. It contains the line $L_1: \frac{x}{3}=\frac{y}{2}=\frac{z}{4}$,so its normal vector $\vec{n}_1$ is perpendicular to $\vec{v}_1 = (3, 2, 4)$.
Let $P_2$ be the plane containing $L_2: \frac{x}{4}=\frac{y}{3}=\frac{z}{2}$ and $L_3: \frac{x}{2}=\frac{y}{-4}=\frac{z}{3}$.
The normal vector $\vec{n}_2$ of $P_2$ is $\vec{v}_2 \times \vec{v}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 2 \\ 2 & -4 & 3 \end{vmatrix} = \hat{i}(9+8) - \hat{j}(12-4) + \hat{k}(-16-6) = 17\hat{i} - 8\hat{j} - 22\hat{k}$.
Since $P_1 \perp P_2$,$\vec{n}_1$ is perpendicular to $\vec{n}_2 = (17, -8, -22)$.
Thus,$\vec{n}_1 = \vec{v}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 4 \\ 17 & -8 & -22 \end{vmatrix} = \hat{i}(-44+32) - \hat{j}(-66-68) + \hat{k}(-24-34) = -12\hat{i} + 134\hat{j} - 58\hat{k}$.
Dividing by $-2$,we get the normal vector $(6, -67, 29)$.
The equation of the plane is $6x - 67y + 29z = 0$.

(C) Let the required plane be $P_1$. It contains the line $L_1: \frac{x}{2}=\frac{y}{3}=\frac{z}{4}$,so its normal vector $\vec{n}_1$ is perpendicular to $\vec{v}_1 = (2, 3, 4)$.
Let $P_2$ be the plane containing the lines $L_2: \frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $L_3: \frac{x}{4}=\frac{y}{2}=\frac{z}{3}$.
The normal vector $\vec{n}_2$ of $P_2$ is given by the cross product of the direction vectors of $L_2$ and $L_3$:
$\vec{n}_2 = (3, 4, 2) \times (4, 2, 3) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 2 \\ 4 & 2 & 3 \end{vmatrix} = \hat{i}(12-4) - \hat{j}(9-8) + \hat{k}(6-16) = (8, -1, -10)$.
Thus,the equation of plane $P_2$ is $8x - y - 10z = 0$.
Since $P_1$ is perpendicular to $P_2$,its normal $\vec{n}_1$ is perpendicular to $\vec{n}_2 = (8, -1, -10)$.
Also,$\vec{n}_1$ is perpendicular to $\vec{v}_1 = (2, 3, 4)$.
Therefore,$\vec{n}_1 = \vec{v}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 8 & -1 & -10 \end{vmatrix} = \hat{i}(-30+4) - \hat{j}(-20-32) + \hat{k}(-2-24) = (-26, 52, -26)$.
Dividing by $-26$,we get the normal vector $(1, -2, 1)$.
The equation of the plane is $1(x-0) - 2(y-0) + 1(z-0) = 0$,which is $x - 2y + z = 0$.

(C) The normal vectors to the given planes are $\vec{n}_1 = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{n}_2 = \hat{i} + 3\hat{j} + 2\hat{k}$.
Since the line $L$ is the intersection of these two planes,it is perpendicular to both normal vectors.
Therefore,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 $\vec{u} = \hat{i} - \hat{j} + \hat{k}$.
The angle $\alpha$ that the line makes with the positive $X$-axis (represented by the unit vector $\hat{i}$) is given by $\cos \alpha = \frac{\vec{u} \cdot \hat{i}}{|\vec{u}| |\hat{i}|}$.
Calculating the dot product: $\vec{u} \cdot \hat{i} = (1)(1) + (-1)(0) + (1)(0) = 1$.
Calculating the magnitude: $|\vec{u}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
Thus,$\cos \alpha = \frac{1}{\sqrt{3} \times 1} = \frac{1}{\sqrt{3}}$.

(A) The equation of a plane passing through a point $\vec{a}$ and parallel to two lines with direction vectors $\vec{b}$ and $\vec{c}$ is given by $(\vec{r} - \vec{a}) \cdot (\vec{b} \times \vec{c}) = 0$,or $\vec{r} \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot (\vec{b} \times \vec{c})$.
Here,the point is $\vec{a} = 2\hat{i} - \hat{j} - 3\hat{k}$ and the direction vectors of the lines are $\vec{b} = 3\hat{i} + 2\hat{j} - 4\hat{k}$ and $\vec{c} = 2\hat{i} - 3\hat{j} + 2\hat{k}$.
First,calculate the normal vector $\vec{n} = \vec{b} \times \vec{c}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -4 \\ 2 & -3 & 2 \end{vmatrix} = \hat{i}(4 - 12) - \hat{j}(6 - (-8)) + \hat{k}(-9 - 4) = -8\hat{i} - 14\hat{j} - 13\hat{k}$.
Now,calculate $\vec{a} \cdot \vec{n}$:
$\vec{a} \cdot \vec{n} = (2)(-8) + (-1)(-14) + (-3)(-13) = -16 + 14 + 39 = 37$.
The equation of the plane is $\vec{r} \cdot (-8\hat{i} - 14\hat{j} - 13\hat{k}) = 37$,which can be written as $-8x - 14y - 13z = 37$,or $8x + 14y + 13z + 37 = 0$.

(B) The equation of the line passing through $(1, -5, 9)$ and parallel to the line $x = y = z$ (direction ratios $1, 1, 1$) 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 $(\lambda + 1, \lambda - 5, \lambda + 9)$.
Since this point lies on the plane $x - y + z = 5$,we substitute these coordinates into the plane equation:
$(\lambda + 1) - (\lambda - 5) + (\lambda + 9) = 5$.
$\lambda + 1 - \lambda + 5 + \lambda + 9 = 5$.
$\lambda + 15 = 5$,which gives $\lambda = -10$.
The point of intersection is $(-10 + 1, -10 - 5, -10 + 9) = (-9, -15, -1)$.
The required distance is the distance between $(1, -5, 9)$ and $(-9, -15, -1)$:
$d = \sqrt{(-9 - 1)^2 + (-15 - (-5))^2 + (-1 - 9)^2}$.
$d = \sqrt{(-10)^2 + (-10)^2 + (-10)^2} = \sqrt{100 + 100 + 100} = \sqrt{300} = 10 \sqrt{3}$ units.

(A) Let $A = (5, -1, 4)$ and $B = (4, -1, 3)$.
The vector $\vec{AB} = (4-5)\hat{i} + (-1-(-1))\hat{j} + (3-4)\hat{k} = -\hat{i} - \hat{k}$.
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 $\cos \phi = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{AB}| |\vec{n}|}$.
$\cos \phi = \frac{|(-1)(1) + (0)(1) + (-1)(1)|}{\sqrt{2} \sqrt{1^2+1^2+1^2}} = \frac{|-2|}{\sqrt{2} \sqrt{3}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}$.
Since $\phi$ is the angle with the normal,the angle $\theta$ with the plane is $90^\circ - \phi$,so $\sin \theta = \cos \phi = \sqrt{\frac{2}{3}}$.
Then $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}}$.
The length of the projection of the line segment on the plane is $|\vec{AB}| \cos \theta = \sqrt{2} \times \sqrt{\frac{1}{3}} = \sqrt{\frac{2}{3}}$ units.

(D) The equation of the plane containing the two given lines is given by the determinant form:
$\begin{vmatrix} x-1 & y-2 & z-3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = 0$
Expanding along the first row:
$(x-1)(15-16) - (y-2)(10-12) + (z-3)(8-9) = 0$
$-(x-1) + 2(y-2) - (z-3) = 0$
$-x + 1 + 2y - 4 - z + 3 = 0$
$-x + 2y - z = 0 \implies x - 2y + z = 0$ ... $(i)$
Given the plane equation $Ax - 2y + z = d$ ... $(ii)$
Since the planes are parallel,the coefficients of $x, y, z$ must be proportional. Thus,$A = 1$.
The distance $D$ between two parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is given by $D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$.
Here,$D = \sqrt{6}$,$a=1, b=-2, c=1$,$d_1 = 0$,and $d_2 = d$.
$\sqrt{6} = \frac{|0 - d|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|d|}{\sqrt{6}}$
$|d| = \sqrt{6} \times \sqrt{6} = 6$.

(B) 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 the plane is parallel to the $Y$-axis,its normal vector must be perpendicular to the $Y$-axis (which has direction vector $\vec{j} = (0, 1, 0)$).
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:
$(1 + 2(-\frac{1}{3}))x + (1 + 3(-\frac{1}{3}))y + (1 - (-\frac{1}{3}))z + (4(-\frac{1}{3}) - 1) = 0$
$(1 - \frac{2}{3})x + 0y + (1 + \frac{1}{3})z + (-\frac{4}{3} - 1) = 0$
$\frac{1}{3}x + \frac{4}{3}z - \frac{7}{3} = 0$
Multiplying by $3$,we get $x+4z-7=0$.

(D) The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Given $a = 1$ and $b = 1$,the equation becomes $x + y + \frac{z}{c} = 1$.
Since the plane passes through $(-1, 1, 2)$,we substitute these coordinates: $-1 + 1 + \frac{2}{c} = 1$,which gives $\frac{2}{c} = 1$,so $c = 2$.
The equation of the plane is $x + y + \frac{z}{2} = 1$,or $2x + 2y + z - 2 = 0$.
The normal vector to the plane is $\vec{n} = 2\hat{i} + 2\hat{j} + 1\hat{k}$.
The direction vector of the $X$-axis is $\vec{v} = 1\hat{i} + 0\hat{j} + 0\hat{k}$.
The angle $\theta$ between a line with direction vector $\vec{v}$ and a plane with normal $\vec{n}$ is given by $\sin \theta = \frac{|\vec{n} \cdot \vec{v}|}{|\vec{n}| |\vec{v}|}$.
$\sin \theta = \frac{|(2)(1) + (2)(0) + (1)(0)|}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{1^2 + 0^2 + 0^2}} = \frac{2}{\sqrt{4 + 4 + 1} \cdot 1} = \frac{2}{\sqrt{9}} = \frac{2}{3}$.
Thus,$\theta = \sin^{-1}\left(\frac{2}{3}\right)$.

(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) 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 $x - 2y + z = 0$.

(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,$a + b + c = 4 + 2 + 3 = 9$.

(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 the plane is parallel to the $X$-axis,its normal vector must be perpendicular to the $X$-axis (direction vector $\vec{i} = (1, 0, 0)$).
Therefore,the coefficient of $x$ must be zero: $1+2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.
Substituting $\lambda = -\frac{1}{2}$ into the equation:
$(1+2(-\frac{1}{2}))x + (1+3(-\frac{1}{2}))y + (1-(-\frac{1}{2}))z + (4(-\frac{1}{2})-1) = 0$
$0x + (1-\frac{3}{2})y + (1+\frac{1}{2})z + (-2-1) = 0$
$-\frac{1}{2}y + \frac{3}{2}z - 3 = 0$
Multiplying by $-2$,we get $y-3z+6=0$.

(D) The equation of any plane passing through the intersection of the planes $P_1: x+2y-3z+1=0$ and $P_2: 3x-2y+4z+3=0$ is given by $P_1 + \lambda P_2 = 0$.
$(x+2y-3z+1) + \lambda(3x-2y+4z+3) = 0 \quad \dots(1)$
Since the plane passes through the point $(2,2,1)$,we substitute $x=2, y=2, z=1$ into equation $(1)$:
$(2 + 2(2) - 3(1) + 1) + \lambda(3(2) - 2(2) + 4(1) + 3) = 0$
$(2 + 4 - 3 + 1) + \lambda(6 - 4 + 4 + 3) = 0$
$4 + 9\lambda = 0$
$\lambda = -\frac{4}{9}$
Substituting $\lambda = -\frac{4}{9}$ back into equation $(1)$:
$(x+2y-3z+1) - \frac{4}{9}(3x-2y+4z+3) = 0$
$9(x+2y-3z+1) - 4(3x-2y+4z+3) = 0$
$9x + 18y - 27z + 9 - 12x + 8y - 16z - 12 = 0$
$-3x + 26y - 43z - 3 = 0$
Multiplying by $-1$,we get $3x - 26y + 43z + 3 = 0$.

(B) Let the direction ratios of the line be $(a, b, c)$. Since the line is parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$,its direction vector is perpendicular to the normals of both planes.
The normals are $\vec{n_1} = (1, -1, 2)$ and $\vec{n_2} = (3, 1, 1)$.
The direction vector $\vec{v}$ is given by the cross product $\vec{n_1} \times \vec{n_2}$:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix} = \hat{i}(-1 - 2) - \hat{j}(1 - 6) + \hat{k}(1 - (-3)) = -3\hat{i} + 5\hat{j} + 4\hat{k}$.
Thus,the direction ratios are $(-3, 5, 4)$.
The line passes through $(1, 2, 3)$,so the Cartesian equation is $\frac{x-1}{-3} = \frac{y-2}{5} = \frac{z-3}{4}$.

(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 the plane is parallel to the $x$-axis,its normal vector $\vec{n} = (1+2\lambda, 1+3\lambda, 1-\lambda)$ must be perpendicular to the $x$-axis direction vector $\vec{i} = (1, 0, 0)$.
Therefore,the dot product is zero: $(1+2\lambda)(1) + (1+3\lambda)(0) + (1-\lambda)(0) = 0$.
$1+2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.
Substituting $\lambda = -\frac{1}{2}$ into the equation:
$(1+2(-\frac{1}{2}))x + (1+3(-\frac{1}{2}))y + (1-(-\frac{1}{2}))z + (4(-\frac{1}{2})-1) = 0$.
$0x + (1-\frac{3}{2})y + (1+\frac{1}{2})z + (-2-1) = 0$.
$-\frac{1}{2}y + \frac{3}{2}z - 3 = 0$.
Multiplying by $-2$,we get $y - 3z + 6 = 0$.

(B) The line passes through the points $A(4, -1, 2)$ and $B(-3, 2, 3)$. Since the line is perpendicular to the plane,the direction ratios of the line are the direction ratios of the normal to the plane.
Direction ratios of the line are $\langle 4 - (-3), -1 - 2, 2 - 3 \rangle = \langle 7, -3, -1 \rangle$.
The plane passes through the point $(-10, 5, 4)$.
The equation of a plane passing through $(x_0, y_0, z_0)$ with normal vector $\langle a, b, c \rangle$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
Substituting the values,we get:
$7(x - (-10)) - 3(y - 5) - 1(z - 4) = 0$
$7(x + 10) - 3(y - 5) - 1(z - 4) = 0$
$7x + 70 - 3y + 15 - z + 4 = 0$
$7x - 3y - z + 89 = 0$.

(B) The equation of the family of planes 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 $(1+2\lambda)x + (2+\lambda)y + (3-\lambda)z + (-4+5\lambda) = 0$ ... $(1)$.
Since this plane is perpendicular to the plane $5x+3y-6z+8=0$,the dot product of their normal vectors must be zero.
Thus,$(1+2\lambda)(5) + (2+\lambda)(3) + (3-\lambda)(-6) = 0$.
Expanding this,we get $5 + 10\lambda + 6 + 3\lambda - 18 + 6\lambda = 0$.
Simplifying,$19\lambda - 7 = 0$,which gives $\lambda = \frac{7}{19}$.
Substituting $\lambda = \frac{7}{19}$ into equation $(1)$:
$(1 + 2(\frac{7}{19}))x + (2 + \frac{7}{19})y + (3 - \frac{7}{19})z + (-4 + 5(\frac{7}{19})) = 0$.
$(\frac{19+14}{19})x + (\frac{38+7}{19})y + (\frac{57-7}{19})z + (\frac{-76+35}{19}) = 0$.
$\frac{33}{19}x + \frac{45}{19}y + \frac{50}{19}z - \frac{41}{19} = 0$.
Multiplying by $19$,we get $33x + 45y + 50z - 41 = 0$.

(B) Let the equation of the plane be $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$. Since the plane passes through the point $(-1, 3, -2)$ (from the line equation),we have $a(x+1) + b(y-3) + c(z+2) = 0$.
Since the plane also passes through $(0, 7, -7)$,we substitute these coordinates: $a(0+1) + b(7-3) + c(-7+2) = 0$,which simplifies to $a + 4b - 5c = 0$.
Also,the line lies in the plane,so the normal vector $(a, b, c)$ is perpendicular to the direction vector of the line $(-3, 2, 1)$. Thus,$-3a + 2b + c = 0$.
Solving the system of equations $a + 4b - 5c = 0$ and $-3a + 2b + c = 0$ using cross multiplication:
$\frac{a}{(4)(1) - (-5)(2)} = \frac{-b}{(1)(1) - (-5)(-3)} = \frac{c}{(1)(2) - (4)(-3)}$
$\frac{a}{4+10} = \frac{-b}{1-15} = \frac{c}{2+12} \Rightarrow \frac{a}{14} = \frac{b}{14} = \frac{c}{14}$.
Taking $a=1, b=1, c=1$,the equation becomes $1(x+1) + 1(y-3) + 1(z+2) = 0$,which simplifies to $x+y+z=0$.

(C) The equation of a plane passing through the intersection of two planes $P_1: x+2y-3z+2=0$ and $P_2: 6x+y+z+1=0$ is given by $P_1 + \lambda P_2 = 0$.
$(x+2y-3z+2) + \lambda(6x+y+z+1) = 0$
$(1+6\lambda)x + (2+\lambda)y + (-3+\lambda)z + (2+\lambda) = 0$
This plane is parallel to the line $\frac{x-1}{1} = \frac{y+2}{1} = \frac{z-7}{-1}$.
The normal vector of the plane is $\vec{n} = (1+6\lambda, 2+\lambda, -3+\lambda)$ and the direction vector of the line is $\vec{v} = (1, 1, -1)$.
Since the plane is parallel to the line,the normal vector is perpendicular to the direction vector,so $\vec{n} \cdot \vec{v} = 0$.
$(1+6\lambda)(1) + (2+\lambda)(1) + (-3+\lambda)(-1) = 0$
$1 + 6\lambda + 2 + \lambda + 3 - \lambda = 0$
$6\lambda + 6 = 0 \Rightarrow \lambda = -1$.
Substituting $\lambda = -1$ into the equation:
$(1-6)x + (2-1)y + (-3-1)z + (2-1) = 0$
$-5x + y - 4z + 1 = 0$
$5x - y + 4z = 1$.

(D) The normal vector $\vec{n}$ of the plane is perpendicular to the direction vectors of both lines,$\vec{v}_1 = 2\hat{i} + 3\hat{j} + 6\hat{k}$ and $\vec{v}_2 = \hat{i} + \hat{j} - \hat{k}$.
Thus,$\vec{n} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 1 & 1 & -1 \end{vmatrix} = \hat{i}(-3-6) - \hat{j}(-2-6) + \hat{k}(2-3) = -9\hat{i} + 8\hat{j} - \hat{k}$.
We can take the normal vector as $\vec{n} = 9\hat{i} - 8\hat{j} + \hat{k}$.
The plane passes through the point $(1, 3, 4)$ from the second line equation.
The equation of the plane is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,where $\vec{a} = \hat{i} + 3\hat{j} + 4\hat{k}$.
$(x\hat{i} + y\hat{j} + z\hat{k} - (\hat{i} + 3\hat{j} + 4\hat{k})) \cdot (9\hat{i} - 8\hat{j} + \hat{k}) = 0$.
$9(x-1) - 8(y-3) + 1(z-4) = 0$.
$9x - 9 - 8y + 24 + z - 4 = 0$.
$9x - 8y + z + 11 = 0$.

(B) The distance of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
Given the points $A(1, 1, \lambda)$ and $B(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0$.
Distance from $A$: $d_1 = \frac{|3(1) + 4(1) - 12(\lambda) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|20 - 12\lambda|}{\sqrt{9 + 16 + 144}} = \frac{|20 - 12\lambda|}{13}$.
Distance from $B$: $d_2 = \frac{|3(-3) + 4(0) - 12(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|-9 + 0 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13}$.
Since $d_1 = d_2$,we have $\frac{|20 - 12\lambda|}{13} = \frac{8}{13}$,which implies $|20 - 12\lambda| = 8$.
Case $1$: $20 - 12\lambda = 8 \Rightarrow 12\lambda = 12 \Rightarrow \lambda = 1$.
Case $2$: $20 - 12\lambda = -8 \Rightarrow 12\lambda = 28 \Rightarrow \lambda = \frac{28}{12} = \frac{7}{3}$.
Since $\lambda$ must be an integer,the value is $\lambda = 1$.

(D) The direction ratios of the given lines are $\vec{b_1} = (1, 4, 7)$ and $\vec{b_2} = (3, 5, 7)$.
Since the plane contains the first line and is parallel to the second,the normal vector $\vec{n}$ to the plane is perpendicular to both $\vec{b_1}$ and $\vec{b_2}$.
$\vec{n} = \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 7 \\ 3 & 5 & 7 \end{vmatrix}$
$= \hat{i}(28-35) - \hat{j}(7-21) + \hat{k}(5-12) = -7\hat{i} + 14\hat{j} - 7\hat{k} = -7(\hat{i} - 2\hat{j} + \hat{k})$.
Thus,the equation of the plane is of the form $x - 2y + z = d$.
Since the plane contains the line,it must pass through the point $(2, 4, 6)$.
Substituting the point into the equation: $2 - 2(4) + 6 = 2 - 8 + 6 = 0$.
Therefore,the equation of the plane is $x - 2y + z = 0$.

(A) The equation of the plane is $\vec{r} \cdot (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) = 13$,which can be written in Cartesian form as $3x + 2y + 6z - 13 = 0$.
Given the point $(x_1, y_1, z_1) = (2, 3, \lambda)$.
The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $ax + by + cz + d = 0$ is given by $d = \left| \frac{ax_1 + by_1 + cz_1 + d}{\sqrt{a^2 + b^2 + c^2}} \right|$.
Substituting the values,we get $5 = \left| \frac{3(2) + 2(3) + 6(\lambda) - 13}{\sqrt{3^2 + 2^2 + 6^2}} \right|$.
$5 = \left| \frac{6 + 6 + 6\lambda - 13}{\sqrt{9 + 4 + 36}} \right|$.
$5 = \left| \frac{6\lambda - 1}{\sqrt{49}} \right|$.
$5 = \left| \frac{6\lambda - 1}{7} \right|$.
$35 = |6\lambda - 1|$.
This gives two cases: $6\lambda - 1 = 35$ or $6\lambda - 1 = -35$.
Case $1$: $6\lambda = 36 \implies \lambda = 6$.
Case $2$: $6\lambda = -34 \implies \lambda = -\frac{34}{6} = -\frac{17}{3}$.
Thus,$\lambda = 6, -\frac{17}{3}$.

(C) The equation of a plane passing through the point $(-1, 3, -2)$ is $a(x+1) + b(y-3) + c(z+2) = 0$.
Since the plane contains the line with direction ratios $(-3, 2, 1)$,we have $-3a + 2b + c = 0$.
The plane also passes through the point $(0, 7, -7)$,so substituting these coordinates into the plane equation gives $a(0+1) + b(7-3) + c(-7+2) = 0$,which simplifies to $a + 4b - 5c = 0$.
Solving the system of equations $-3a + 2b + c = 0$ and $a + 4b - 5c = 0$ using cross-multiplication:
$\frac{a}{2(-5) - 1(4)} = \frac{b}{1(1) - (-3)(-5)} = \frac{c}{-3(4) - 2(1)}$
$\frac{a}{-10-4} = \frac{b}{1-15} = \frac{c}{-12-2}$
$\frac{a}{-14} = \frac{b}{-14} = \frac{c}{-14} \implies a=1, b=1, c=1$.
Substituting these values into the plane equation: $1(x+1) + 1(y-3) + 1(z+2) = 0 \implies x+y+z=0$.

(C) The line is given by $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$,so its direction vector is $\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}$.
The plane is given by $2x - y + \sqrt{\lambda}z + 4 = 0$,so its normal vector is $\vec{n} = 2\hat{i} - \hat{j} + \sqrt{\lambda}\hat{k}$.
The angle $\theta$ between a line with direction $\vec{b}$ and a plane with normal $\vec{n}$ is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Given $\sin \theta = \frac{1}{3}$,we have $\frac{1}{3} = \frac{|(1)(2) + (2)(-1) + (2)(\sqrt{\lambda})|}{\sqrt{1^2 + 2^2 + 2^2} \sqrt{2^2 + (-1)^2 + (\sqrt{\lambda})^2}}$.
$\frac{1}{3} = \frac{|2 - 2 + 2\sqrt{\lambda}|}{\sqrt{9} \sqrt{5 + \lambda}} = \frac{2\sqrt{\lambda}}{3\sqrt{5 + \lambda}}$.
Squaring both sides: $\frac{1}{9} = \frac{4\lambda}{9(5 + \lambda)}$.
$5 + \lambda = 4\lambda$,which implies $3\lambda = 5$,so $\lambda = \frac{5}{3}$.
Therefore,$\lambda + 1 = \frac{5}{3} + 1 = \frac{8}{3}$.

(D) The line of intersection of two planes is perpendicular to the normal vectors of both planes. Let the normal vectors be $\vec{n}_1 = 3 \hat{i}-\hat{j}+\hat{k}$ and $\vec{n}_2 = \hat{i}+4 \hat{j}-2 \hat{k}$.
The direction vector $\vec{v}$ of the line of intersection is given by the cross product of the normal vectors: $\vec{v} = \vec{n}_1 \times \vec{n}_2$.
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{vmatrix}$.
$\vec{v} = \hat{i}((-1)(-2) - (1)(4)) - \hat{j}((3)(-2) - (1)(1)) + \hat{k}((3)(4) - (-1)(1))$.
$\vec{v} = \hat{i}(2 - 4) - \hat{j}(-6 - 1) + \hat{k}(12 + 1)$.
$\vec{v} = -2 \hat{i} + 7 \hat{j} + 13 \hat{k}$.
Thus,the line is parallel to the vector $-2 \hat{i} + 7 \hat{j} + 13 \hat{k}$.

(B) The lines are given by $L_1: \frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $L_2: \frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$.
Since the lines are coplanar,the condition is $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$.
Here,$(x_1, y_1, z_1) = (1, -1, 0)$ and $(x_2, y_2, z_2) = (-1, -1, 0)$.
So,$\begin{vmatrix} -1-1 & -1-(-1) & 0-0 \\ 2 & k & 2 \\ 5 & 2 & k \end{vmatrix} = 0 \implies \begin{vmatrix} -2 & 0 & 0 \\ 2 & k & 2 \\ 5 & 2 & k \end{vmatrix} = 0$.
Expanding along the first row: $-2(k^2 - 4) = 0 \implies k^2 = 4 \implies k = \pm 2$.
Case $1$: If $k=2$,the lines are $\frac{x-1}{2}=\frac{y+1}{2}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{2}$. The normal vector is $\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 2 \\ 5 & 2 & 2 \end{vmatrix} = 0\hat{i} + 6\hat{j} - 6\hat{k}$. The plane equation is $0(x-1) + 6(y+1) - 6(z-0) = 0 \implies y - z + 1 = 0$.
Case $2$: If $k=-2$,the lines are $\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{-2}$. The normal vector is $\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 2 \\ 5 & 2 & -2 \end{vmatrix} = 0\hat{i} + 14\hat{j} + 14\hat{k}$. The plane equation is $0(x-1) + 14(y+1) + 14(z-0) = 0 \implies y + z + 1 = 0$.
Combining both,the equation is $y \pm z + 1 = 0$.

(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 $\theta = \cos^{-1} \sqrt{\frac{5}{14}}$,we have $\cos \theta = \sqrt{\frac{5}{14}}$,so $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{5}{14} = \frac{9}{14}$.
Thus,$\sin \theta = \frac{3}{\sqrt{14}}$.
The formula for $\sin \theta$ is $\frac{|(1)(1) + (2)(2) + (\lambda)(3)|}{\sqrt{1^2 + 2^2 + \lambda^2} \sqrt{1^2 + 2^2 + 3^2}} = \frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2} \sqrt{14}}$.
Equating the two expressions for $\sin \theta$: $\frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2} \sqrt{14}} = \frac{3}{\sqrt{14}}$.
$|5 + 3\lambda| = 3\sqrt{5 + \lambda^2}$.
Squaring both sides: $(5 + 3\lambda)^2 = 9(5 + \lambda^2)$.
$25 + 30\lambda + 9\lambda^2 = 45 + 9\lambda^2$.
$30\lambda = 20$.
$\lambda = \frac{20}{30} = \frac{2}{3}$.

(A) The equation of the line passing through $Q(3,-4,-5)$ and $R(2,-3,1)$ is given by $\frac{x-3}{2-3} = \frac{y-(-4)}{-3-(-4)} = \frac{z-(-5)}{1-(-5)}$.
This simplifies to $\frac{x-3}{-1} = \frac{y+4}{1} = \frac{z+5}{6} = k$.
Any point on this line is $(3-k, -4+k, -5+6k)$.
Since this point lies on the plane $2x+y+z=7$,we substitute the coordinates into the plane equation:
$2(3-k) + (-4+k) + (-5+6k) = 7$.
$6 - 2k - 4 + k - 5 + 6k = 7$.
$5k - 3 = 7 \implies 5k = 10 \implies k = 2$.
The point of intersection is $(3-2, -4+2, -5+12) = (1, -2, 7)$.
The distance between $P(3,4,4)$ and $(1,-2,7)$ is $\sqrt{(3-1)^2 + (4-(-2))^2 + (4-7)^2}$.
$= \sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$ units.

(A) The lines are $L_1: \frac{x}{1} = \frac{y-2}{2} = \frac{z+3}{\lambda}$ and $L_2: \frac{x-2}{2} = \frac{y-6}{3} = \frac{z-3}{\lambda}$.
Points on the lines are $A(0, 2, -3)$ and $B(2, 6, 3)$.
Direction vectors are $\vec{v_1} = \hat{i} + 2\hat{j} + \lambda\hat{k}$ and $\vec{v_2} = 2\hat{i} + 3\hat{j} + \lambda\hat{k}$.
The normal to the plane is $\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & \lambda \\ 2 & 3 & \lambda \end{vmatrix} = \hat{i}(2\lambda - 3\lambda) - \hat{j}(\lambda - 2\lambda) + \hat{k}(3 - 4) = -\lambda\hat{i} + \lambda\hat{j} - \hat{k}$.
The equation of the plane is $-\lambda(x - 0) + \lambda(y - 2) - 1(z + 3) = 0$,which simplifies to $-\lambda x + \lambda y - 2\lambda - z - 3 = 0$.
Rearranging,we get $-\lambda(x - y + 2) - (z + 3) = 0$.
For this to hold for all $\lambda$,we must have $x - y + 2 = 0$ and $z + 3 = 0$.
Thus,the plane always passes through the line of intersection of $x - y + 2 = 0$ and $z = -3$.
Checking the options: For $(2, 4, -3)$,$2 - 4 + 2 = 0$ and $-3 = -3$. None of the given options satisfy this directly,but re-evaluating the point $(2, 4, -3)$ is not listed. Checking $(4, 10, 9)$ is incorrect. Let's re-check the point $(2, 8, 7)$ or others. Actually,the point $(2, 4, -3)$ is the intersection. Testing $(4, 6, -3)$ or similar. Given the options,let's test $(2, 4, -3)$ which is not there. Re-checking the question,the correct point is $(2, 4, -3)$. Since it is not in the options,we identify the point that satisfies the plane equation for all $\lambda$.

(D) The line passes through the point $P(3, -5, -2)$. Since the line lies in the plane $\alpha x + 3y - z + \beta = 0$,the point $P$ must satisfy the plane equation:
$\alpha(3) + 3(-5) - (-2) + \beta = 0$
$3\alpha - 15 + 2 + \beta = 0$
$3\alpha + \beta = 13$ (Equation $1$).
Also,the direction vector of the line $\vec{v} = 2\hat{i} - \hat{j} + 2\hat{k}$ must be perpendicular to the normal vector of the plane $\vec{n} = \alpha\hat{i} + 3\hat{j} - \hat{k}$.
Thus,$\vec{v} \cdot \vec{n} = 0$:
$(2)(\alpha) + (-1)(3) + (2)(-1) = 0$
$2\alpha - 3 - 2 = 0$
$2\alpha = 5$
$\alpha = \frac{5}{2}$.
Substituting $\alpha = \frac{5}{2}$ into Equation $1$:
$3(\frac{5}{2}) + \beta = 13$
$\frac{15}{2} + \beta = 13$
$\beta = 13 - \frac{15}{2} = \frac{26-15}{2} = \frac{11}{2}$.
Therefore,the values are $\alpha = \frac{5}{2}$ and $\beta = \frac{11}{2}$.

(D) First,rewrite the lines in symmetric form:
Line $1$: $x = a(y - \frac{1}{a}) = z - 2 \implies \frac{x}{1} = \frac{y - 1/a}{1/a} = \frac{z - 2}{1}$. Point $P_1 = (0, 1/a, 2)$,direction vector $\vec{v_1} = (1, 1/a, 1)$.
Line $2$: $x = 3(y - 2/3) = b(z - 2/b) \implies \frac{x}{1} = \frac{y - 2/3}{1/3} = \frac{z - 2/b}{1/b}$. Point $P_2 = (0, 2/3, 2/b)$,direction vector $\vec{v_2} = (1, 1/3, 1/b)$.
For lines to be coplanar,the scalar triple product $[\vec{P_1P_2}, \vec{v_1}, \vec{v_2}] = 0$.
$\vec{P_1P_2} = (0, 2/3 - 1/a, 2/b - 2)$.
The determinant is:
$|\begin{matrix} 0 & 2/3 - 1/a & 2/b - 2 \\ 1 & 1/a & 1 \\ 1 & 1/3 & 1/b \end{matrix}| = 0$.
Expanding along the first row:
$-(2/3 - 1/a)(1/b - 1) + (2/b - 2)(1/3 - 1/a) = 0$.
$-(2/3 - 1/a)(1/b - 1) + 2(1/b - 1)(1/3 - 1/a) = 0$.
$(1/b - 1) [-(2/3 - 1/a) + 2(1/3 - 1/a)] = 0$.
$(1/b - 1) [-2/3 + 1/a + 2/3 - 2/a] = 0$.
$(1/b - 1)(-1/a) = 0$.
Since $a \neq 0$,we must have $1/b - 1 = 0$,which implies $b = 1$.
Thus,$b = 1$ and $a$ can be any non-zero real number. The correct option is $D$.

(A) The equation of the line passing through $A(3, -4, 5)$ and parallel to the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{-2}$ is given by $\frac{x-3}{2} = \frac{y+4}{1} = \frac{z-5}{-2} = r$.
Any point on this line is of the form $P(2r+3, r-4, -2r+5)$.
Since this point $P$ lies on the plane $2x + 5y - 6z = 16$,we substitute the coordinates of $P$ into the plane equation:
$2(2r+3) + 5(r-4) - 6(-2r+5) = 16$.
$4r + 6 + 5r - 20 + 12r - 30 = 16$.
$21r - 44 = 16$.
$21r = 60$.
$r = \frac{60}{21} = \frac{20}{7}$.
The distance $AP$ is the distance between $(3, -4, 5)$ and $(2r+3, r-4, -2r+5)$,which is $\sqrt{(2r)^2 + (r)^2 + (-2r)^2} = \sqrt{4r^2 + r^2 + 4r^2} = \sqrt{9r^2} = 3|r|$.
Substituting $r = \frac{20}{7}$,we get $3 \times \frac{20}{7} = \frac{60}{7}$ units.

(D) The equation of the line passing through $(1, 1, 1)$ and $(2, 2, 2)$ is given by $\frac{x-1}{2-1} = \frac{y-1}{2-1} = \frac{z-1}{2-1} = k$.
This simplifies to $x-1 = k$,$y-1 = k$,and $z-1 = k$.
So,any point on the line is of the form $(k+1, k+1, k+1)$.
Since this point lies on the plane $x + y + z = 9$,we substitute these coordinates into the plane equation:
$(k+1) + (k+1) + (k+1) = 9$.
$3k + 3 = 9$.
$3k = 6$,which gives $k = 2$.
Substituting $k = 2$ back into the point coordinates $(k+1, k+1, k+1)$,we get $(2+1, 2+1, 2+1) = (3, 3, 3)$.
Thus,the point of intersection is $(3, 3, 3)$.

(C) Since the line lies in the plane,the direction vector of the line is perpendicular to the normal vector of the plane. The direction vector of the line is $\vec{v} = 2\hat{i} - \hat{j} + 3\hat{k}$ and the normal to the plane is $\vec{n} = \ell\hat{i} + m\hat{j} - \hat{k}$.
Thus,$\vec{v} \cdot \vec{n} = 0 \Rightarrow (2)(\ell) + (-1)(m) + (3)(-1) = 0 \Rightarrow 2\ell - m = 3$ (Equation $i$).
Also,any point on the line must satisfy the plane equation. The point $(3, -2, -4)$ lies on the line,so it must lie on the plane:
$\ell(3) + m(-2) - (-4) = 9 \Rightarrow 3\ell - 2m + 4 = 9 \Rightarrow 3\ell - 2m = 5$ (Equation $ii$).
Solving the system of equations:
From $(i)$,$m = 2\ell - 3$. Substituting into $(ii)$:
$3\ell - 2(2\ell - 3) = 5 \Rightarrow 3\ell - 4\ell + 6 = 5 \Rightarrow -\ell = -1 \Rightarrow \ell = 1$.
Then $m = 2(1) - 3 = -1$.
Therefore,$\ell^2 + m^2 = (1)^2 + (-1)^2 = 1 + 1 = 2$.

(D) Let the given line be $L: \frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5} = k$. Any point on $L$ is $(3k+1, k+3, -5k+4)$.
For the intersection point of $L$ and the plane $2x-y+z+3=0$,we have $2(3k+1)-(k+3)+(-5k+4)+3=0$.
$6k+2-k-3-5k+4+3=0 \implies 6=0$,which is impossible. Thus,the line is parallel to the plane.
The image of point $P(1, 3, 4)$ on the line in the plane $2x-y+z+3=0$ is $P'(x', y', z')$.
Using the formula $\frac{x'-1}{2} = \frac{y'-3}{-1} = \frac{z'-4}{1} = -2 \frac{2(1)-3+4+3}{2^2+(-1)^2+1^2} = -2 \frac{6}{6} = -2$.
$x'-1 = -4 \implies x' = -3$.
$y'-3 = 2 \implies y' = 5$.
$z'-4 = -2 \implies z' = 2$.
The image line passes through $(-3, 5, 2)$ and has the same direction ratios $(3, 1, -5)$ as the original line.
Thus,the equation is $\frac{x+3}{3} = \frac{y-5}{1} = \frac{z-2}{-5}$.

(C) The given line is $\frac{x-4}{1} = \frac{y-2}{1} = \frac{z - m/2}{3/2}$.
This line passes through the point $P(4, 2, m/2)$.
Since the line lies in the plane $2x - 5y + 2z = 7$,the point $P$ must satisfy the equation of the plane.
Substituting the coordinates of $P$ into the plane equation:
$2(4) - 5(2) + 2(m/2) = 7$
$8 - 10 + m = 7$
$-2 + m = 7$
$m = 9$
Also,the direction vector of the line $\vec{v} = (1, 1, 3/2)$ must be perpendicular to the normal of the plane $\vec{n} = (2, -5, 2)$.
Checking the dot product: $\vec{v} \cdot \vec{n} = (1)(2) + (1)(-5) + (3/2)(2) = 2 - 5 + 3 = 0$.
Since the dot product is $0$,the line is parallel to the plane,and since the point $P$ lies on the plane,the entire line lies in the plane for $m = 9$.

(A) Let the line be $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}=\lambda$.
Any point on this line is given by $(2\lambda+4, 2\lambda+5, \lambda+3)$.
Since this point lies on the plane $x+y+z=2$,we substitute the coordinates into the plane equation:
$(2\lambda+4) + (2\lambda+5) + (\lambda+3) = 2$.
$5\lambda + 12 = 2$.
$5\lambda = -10$,so $\lambda = -2$.
Substituting $\lambda = -2$ back into the point coordinates:
$x = 2(-2) + 4 = 0$,
$y = 2(-2) + 5 = 1$,
$z = (-2) + 3 = 1$.
The point of intersection is $(0, 1, 1)$.
Now,check which option satisfies the point $(0, 1, 1)$:
For option $(A)$: $\frac{0-1}{1} = -1$,$\frac{1-3}{2} = -1$,$\frac{1+4}{-5} = -1$.
Since all ratios are equal to $-1$,the point $(0, 1, 1)$ lies on the line given in option $(A)$.