Line Questions in English

(A) Let the direction ratios of the two lines be $\vec{b_1} = (1, 2, 2)$ and $\vec{b_2} = (2, 2, -1)$.
The formula for the angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by:
$\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$
Substituting the values:
$\cos \theta = \left| \frac{1(2) + 2(2) + 2(-1)}{\sqrt{1^2 + 2^2 + 2^2} \sqrt{2^2 + 2^2 + (-1)^2}} \right|$
$\cos \theta = \left| \frac{2 + 4 - 2}{\sqrt{1 + 4 + 4} \sqrt{4 + 4 + 1}} \right|$
$\cos \theta = \left| \frac{4}{\sqrt{9} \sqrt{9}} \right| = \frac{4}{3 \times 3} = \frac{4}{9}$
Therefore,$\theta = \cos ^{-1}\left(\frac{4}{9}\right)$.

(A) The direction vectors of the two lines are $\vec{b_1} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b_2} = 2\hat{i} + \hat{j} - \hat{k}$.
Let $\theta$ be the angle between the lines.
The formula for the angle between two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ is given by $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.
Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(2) + (1)(1) + (2)(-1) = 2 + 1 - 2 = 1$.
Calculating the magnitudes: $|\vec{b_1}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$ and $|\vec{b_2}| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{6}$.
Substituting these values into the formula: $\cos \theta = \frac{|1|}{\sqrt{6} \cdot \sqrt{6}} = \frac{1}{6}$.
Therefore,$\theta = \cos^{-1}\left(\frac{1}{6}\right)$.

(D) Let $\vec{a}$ and $\vec{b}$ be the direction vectors of the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$ respectively.
$\vec{a} = 4\hat{i} + \hat{j} + 8\hat{k}$ and $\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}$.
The dot product is $\vec{a} \cdot \vec{b} = (4 \times 2) + (1 \times 2) + (8 \times 1) = 8 + 2 + 8 = 18$.
The magnitudes are $|\vec{a}| = \sqrt{4^2 + 1^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9$ and $|\vec{b}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Let $\theta$ be the angle between the lines. Then $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{18}{9 \times 3} = \frac{18}{27} = \frac{2}{3}$.
Therefore,$\theta = \cos^{-1}\left(\frac{2}{3}\right)$.

(B) The direction vectors of the two lines are $\vec{b_1} = \hat{\imath} - 2\hat{\jmath} - 2\hat{k}$ and $\vec{b_2} = 3\hat{\imath} + 2\hat{\jmath} - 6\hat{k}$.
The cosine of the angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.
First,calculate the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(3) + (-2)(2) + (-2)(-6) = 3 - 4 + 12 = 11$.
Next,calculate the magnitudes: $|\vec{b_1}| = \sqrt{1^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$|\vec{b_2}| = \sqrt{3^2 + 2^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
Therefore,$\cos \theta = \frac{|11|}{3 \times 7} = \frac{11}{21}$.

(D) The direction vector of the first line is $\vec{b_1} = \hat{i} - \hat{j} + \hat{k}$.
The direction vector of the second line is $\vec{b_2} = \hat{i} + 3\hat{j} + 2\hat{k}$.
The angle $\theta$ between the two lines is given by $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.
Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(1) + (-1)(3) + (1)(2) = 1 - 3 + 2 = 0$.
Since the dot product is $0$,$\cos \theta = 0$,which implies $\theta = \frac{\pi}{2}$.

(A) Let the direction ratios of the required line be $(a, b, c)$.
The direction ratios of the first line are $(1, 2, 4)$ and the second line are $(2, 2, 1)$.
Since the required line is perpendicular to both,its direction vector is the cross product of the direction vectors of the given lines:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 2 & 1 \end{vmatrix} = \hat{i}(2-8) - \hat{j}(1-8) + \hat{k}(2-4) = -6\hat{i} + 7\hat{j} - 2\hat{k}$.
Thus,the direction ratios are proportional to $(-6, 7, -2)$ or $(6, -7, 2)$.
The line passes through $(1, 2, 3)$,so its equation is $\frac{x-1}{6} = \frac{y-2}{-7} = \frac{z-3}{2}$.
This can be rewritten as $\frac{x-1}{6} = \frac{2-y}{7} = \frac{z-3}{2}$.

(C) The vector equation of a line passing through points $A(\vec{a})$ and $B(\vec{b})$ is given by $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$.
Here,$\vec{a} = 3\hat{i} + 4\hat{j} - 7\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 6\hat{k}$.
The direction vector $\vec{v} = \vec{b} - \vec{a} = (1-3)\hat{i} + (-1-4)\hat{j} + (6 - (-7))\hat{k} = -2\hat{i} - 5\hat{j} + 13\hat{k}$.
Thus,the parametric equations are $x = x_1 + v_x \lambda$,$y = y_1 + v_y \lambda$,$z = z_1 + v_z \lambda$.
Substituting the values: $x = 3 - 2\lambda$,$y = 4 - 5\lambda$,$z = -7 + 13\lambda$.

(D) Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Given the direction ratios of the first line are $(-3, 2k, 2)$ and the second line are $(3k, 1, -5)$.
Applying the condition for perpendicularity:
$(-3)(3k) + (2k)(1) + (2)(-5) = 0$
$-9k + 2k - 10 = 0$
$-7k - 10 = 0$
$-7k = 10$
$k = \frac{-10}{7}$

(D) Key Idea: If lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ are perpendicular,then $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Given lines are:
$\frac{2x-4}{\lambda} = \frac{y-1}{2} = \frac{z-3}{1} \Rightarrow \frac{x-2}{\lambda/2} = \frac{y-1}{2} = \frac{z-3}{1}$ $(i)$
And $\frac{x-1}{1} = \frac{3y-1}{\lambda} = \frac{z-2}{1} \Rightarrow \frac{x-1}{1} = \frac{y-1/3}{\lambda/3} = \frac{z-2}{1}$ (ii)
Since lines $(i)$ and (ii) are perpendicular,the dot product of their direction ratios must be zero:
$(\frac{\lambda}{2})(1) + (2)(\frac{\lambda}{3}) + (1)(1) = 0$
$\frac{\lambda}{2} + \frac{2\lambda}{3} + 1 = 0$
Multiplying by $6$ to clear the denominators:
$3\lambda + 4\lambda + 6 = 0$
$7\lambda = -6$
$\lambda = -\frac{6}{7}$

(C) Let the given lines be:
$L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = k_1$
$L_2: \frac{x-3}{1}=\frac{y-\lambda}{2}=\frac{z}{1} = k_2$
Any point on $L_1$ is $(2k_1+1, 3k_1-1, 4k_1+1)$ and any point on $L_2$ is $(k_2+3, 2k_2+\lambda, k_2)$.
If the lines intersect,there exist $k_1, k_2$ such that:
$2k_1+1 = k_2+3 \Rightarrow 2k_1 - k_2 = 2$ $(i)$
$4k_1+1 = k_2 \Rightarrow 4k_1 - k_2 = -1$ $(ii)$
Subtracting $(i)$ from $(ii)$:
$(4k_1 - k_2) - (2k_1 - k_2) = -1 - 2$
$2k_1 = -3 \Rightarrow k_1 = -\frac{3}{2}$
Substituting $k_1$ in $(ii)$:
$4(-\frac{3}{2}) - k_2 = -1 \Rightarrow -6 - k_2 = -1 \Rightarrow k_2 = -5$
Now,equate the $y$-coordinates:
$3k_1 - 1 = 2k_2 + \lambda$
$3(-\frac{3}{2}) - 1 = 2(-5) + \lambda$
$-\frac{9}{2} - 1 = -10 + \lambda$
$-\frac{11}{2} = -10 + \lambda$
$\lambda = 10 - \frac{11}{2} = \frac{20-11}{2} = \frac{9}{2}$

(C) Since the points $P(6, -1, 2)$,$Q(8, -7, 2\lambda)$,and $R(5, 2, 4)$ are collinear,the direction ratios of the line segments $PQ$ and $PR$ must be proportional.
The direction ratios of $PQ$ are $(8-6, -7-(-1), 2\lambda-2) = (2, -6, 2\lambda-2)$.
The direction ratios of $PR$ are $(5-6, 2-(-1), 4-2) = (-1, 3, 2)$.
Since the points are collinear,the ratio of the direction ratios must be equal:
$\frac{2}{-1} = \frac{-6}{3} = \frac{2\lambda-2}{2}$
$-2 = -2 = \lambda-1$
Equating the last part: $-2 = \lambda-1$,which gives $\lambda = -1$.

(D) The direction ratios of the first line are $\vec{a_1} = (2, -2, 1)$.
The direction ratios of the second line are $\vec{a_2} = (1, 2, 2)$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Substituting the values:
$\cos \theta = \frac{|(2)(1) + (-2)(2) + (1)(2)|}{\sqrt{2^2 + (-2)^2 + 1^2} \sqrt{1^2 + 2^2 + 2^2}}$
$\cos \theta = \frac{|2 - 4 + 2|}{\sqrt{4 + 4 + 1} \sqrt{1 + 4 + 4}}$
$\cos \theta = \frac{0}{\sqrt{9} \sqrt{9}} = 0$
Since $\cos \theta = 0$,we have $\theta = 90^{\circ}$.

(C) Let the direction vectors of the two given lines be $\vec{b_1} = 2\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b_2} = \hat{i} - 2\hat{j} + 2\hat{k}$.
Since the required line is perpendicular to both lines,its direction vector $\vec{b}$ is given by the cross product $\vec{b_1} \times \vec{b_2}$.
$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -2 & 2 \end{vmatrix} = \hat{i}(-4 + 2) - \hat{j}(4 - 1) + \hat{k}(-4 + 2) = -2\hat{i} - 3\hat{j} - 2\hat{k}$.
We can take the direction ratios as $(2, 3, 2)$ by multiplying by $-1$.
The line passes through the point $(3, -1, 2)$.
Thus,the equation of the line is $\frac{x - 3}{2} = \frac{y - (-1)}{3} = \frac{z - 2}{2}$,which simplifies to $\frac{x - 3}{2} = \frac{y + 1}{3} = \frac{z - 2}{2}$.

(A) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ intersect if and only if the determinant of the matrix formed by the differences of their points and their direction ratios is zero:
$\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right| = 0$
Given the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z-0}{1}$,we have $(x_1, y_1, z_1) = (1, -1, 1)$ and $(x_2, y_2, z_2) = (3, k, 0)$.
The direction ratios are $(a_1, b_1, c_1) = (2, 3, 4)$ and $(a_2, b_2, c_2) = (1, 2, 1)$.
Substituting these into the determinant condition:
$\left|\begin{array}{ccc} 3-1 & k-(-1) & 0-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right| = 0$
$\Rightarrow \left|\begin{array}{ccc} 2 & k+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right| = 0$
Expanding along the first row:
$2(3(1) - 4(2)) - (k+1)(2(1) - 4(1)) - 1(2(2) - 3(1)) = 0$
$2(3-8) - (k+1)(2-4) - 1(4-3) = 0$
$2(-5) - (k+1)(-2) - 1(1) = 0$
$-10 + 2k + 2 - 1 = 0$
$2k - 9 = 0$
$2k = 9$
$k = \frac{9}{2}$

(B) Let the general point on the first line $L_1$ be $P(r)$.
$\frac{x - 1}{2} = \frac{y + 1}{2} = \frac{z - 1}{4} = r$
$x = 2r + 1, y = 2r - 1, z = 4r + 1$
So,$P = (2r + 1, 2r - 1, 4r + 1)$.
Since the lines intersect,this point must satisfy the equation of the second line $L_2$: $\frac{x - 3}{1} = \frac{y - 6}{2} = \frac{z}{1} = k$.
Substituting the coordinates of $P$ into $L_2$:
$\frac{2r + 1 - 3}{1} = \frac{2r - 1 - 6}{2} = \frac{4r + 1}{1}$
$\frac{2r - 2}{1} = \frac{2r - 7}{2} = 4r + 1$
Taking the first and third parts: $2r - 2 = 4r + 1 \Rightarrow -3 = 2r \Rightarrow r = -\frac{3}{2}$.
Now,substitute $r = -\frac{3}{2}$ into the coordinates of $P$:
$x = 2(-\frac{3}{2}) + 1 = -3 + 1 = -2$
$y = 2(-\frac{3}{2}) - 1 = -3 - 1 = -4$
$z = 4(-\frac{3}{2}) + 1 = -6 + 1 = -5$
Thus,the intersection point is $(-2, -4, -5)$.

(A) The equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Given that the line passes through $(-3, 2, -5)$,we have $x_1 = -3, y_1 = 2, z_1 = -5$.
Since the line is equally inclined to the positive coordinate axes,the direction cosines $(l, m, n)$ are equal,i.e.,$l = m = n$.
Thus,the direction ratios can be taken as $a = 1, b = 1, c = 1$.
Substituting these values into the equation,we get $\frac{x - (-3)}{1} = \frac{y - 2}{1} = \frac{z - (-5)}{1}$.
Therefore,the equation of the line is $\frac{x+3}{1} = \frac{y-2}{1} = \frac{z+5}{1}$.

(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.

(A) The direction vectors of the two given lines are $\vec{b_1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $\vec{b_2} = -3 \hat{i} + 2 \hat{j} + 5 \hat{k}$.
Since the required line is perpendicular to both,its direction vector $\vec{v}$ is given by the cross product $\vec{b_1} \times \vec{b_2}$.
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & 2 & 5 \end{vmatrix} = \hat{i}(10 - 6) - \hat{j}(5 - (-9)) + \hat{k}(2 - (-6)) = 4 \hat{i} - 14 \hat{j} + 8 \hat{k}$.
The line passes through $(1, 2, 3)$,so its vector equation is $\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \lambda(4 \hat{i} - 14 \hat{j} + 8 \hat{k})$.

(A) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are coplanar if $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$.
Here,$(x_1, y_1, z_1) = (3, 2, 5)$,$(a_1, b_1, c_1) = (1, 1, -k)$,$(x_2, y_2, z_2) = (4, 3, 3)$,and $(a_2, b_2, c_2) = (k, 1, 2)$.
The condition becomes $\begin{vmatrix} 4-3 & 3-2 & 3-5 \\ 1 & 1 & -k \\ k & 1 & 2 \end{vmatrix} = 0$.
$\begin{vmatrix} 1 & 1 & -2 \\ 1 & 1 & -k \\ k & 1 & 2 \end{vmatrix} = 0$.
Expanding along the first row: $1(2 - (-k)) - 1(2 - (-k^2)) - 2(1 - k) = 0$.
$1(2 + k) - 1(2 + k^2) - 2 + 2k = 0$.
$2 + k - 2 - k^2 - 2 + 2k = 0$.
$-k^2 + 3k - 2 = 0$.
$k^2 - 3k + 2 = 0$.
$(k-1)(k-2) = 0$.
Thus,$k = 1$ or $k = 2$.

(A) Let $A(x_1, y_1, z_1) = (3, 4, 1)$ and $B(x_2, y_2, z_2) = (5, 1, 6)$.
The equation of the line passing through two points is $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.
Substituting the values,we get $\frac{x-3}{5-3} = \frac{y-4}{1-4} = \frac{z-1}{6-1}$,which simplifies to $\frac{x-3}{2} = \frac{y-4}{-3} = \frac{z-1}{5} = k$.
Since the line crosses the $xy$-plane,the $z$-coordinate must be $0$.
Setting $z = 0$,we have $\frac{0-1}{5} = k$,so $k = -\frac{1}{5}$.
Now,find $x$ and $y$ using $k = -\frac{1}{5}$:
$x - 3 = 2k \Rightarrow x = 3 + 2(-\frac{1}{5}) = 3 - \frac{2}{5} = \frac{13}{5}$.
$y - 4 = -3k \Rightarrow y = 4 - 3(-\frac{1}{5}) = 4 + \frac{3}{5} = \frac{23}{5}$.
Thus,the required point is $\left(\frac{13}{5}, \frac{23}{5}, 0\right)$.

(B) The lines $L_1$ and $L_2$ are coplanar if the determinant of the vector connecting a point on each line and their direction vectors is zero.
Given points: $P_1 = (-1, 2, 1)$ and $P_2 = (-2, -1, -1)$.
Direction vectors: $\vec{v_1} = (2, -1, 1)$ and $\vec{v_2} = (\alpha, 5-\alpha, 1)$.
The condition for coplanarity is:
$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$
$\begin{vmatrix} -2-(-1) & -1-2 & -1-1 \\ 2 & -1 & 1 \\ \alpha & 5-\alpha & 1 \end{vmatrix} = 0$
$\begin{vmatrix} -1 & -3 & -2 \\ 2 & -1 & 1 \\ \alpha & 5-\alpha & 1 \end{vmatrix} = 0$
Expanding the determinant:
$-1(-1 - (5-\alpha)) + 3(2 - \alpha) - 2(2(5-\alpha) - (-1)\alpha) = 0$
$-1(-6+\alpha) + 6 - 3\alpha - 2(10 - 2\alpha + \alpha) = 0$
$6 - \alpha + 6 - 3\alpha - 20 + 2\alpha = 0$
$-2\alpha - 8 = 0 \Rightarrow \alpha = -4$.
Substituting $\alpha = -4$ into $L_2$:
$L_2: \frac{x+2}{-4} = \frac{y+1}{5-(-4)} = \frac{z+1}{1} \Rightarrow \frac{x+2}{-4} = \frac{y+1}{9} = \frac{z+1}{1}$.
Checking option $(B) (2, -10, -2)$:
$\frac{2+2}{-4} = \frac{-10+1}{9} = \frac{-2+1}{1} \Rightarrow -1 = -1 = -1$.
Thus,the line $L_2$ passes through $(2, -10, -2)$.

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

(D) Let the given lines be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$.
Any point on the first line is $(2\lambda+1, 3\lambda-1, 4\lambda+1)$ and any point on the second line is $(\mu+3, 2\mu+k, \mu)$.
Since the lines intersect,there must exist some $\lambda$ and $\mu$ such that the coordinates are equal:
$2\lambda+1 = \mu+3 \implies 2\lambda - \mu = 2$ .... $(1)$
$3\lambda-1 = 2\mu+k \implies 3\lambda - 2\mu = k+1$ .... $(2)$
$4\lambda+1 = \mu \implies 4\lambda - \mu = -1$ .... $(3)$
Subtracting $(1)$ from $(3)$,we get $(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2$,which gives $2\lambda = -3$,so $\lambda = \frac{-3}{2}$.
Substituting $\lambda = \frac{-3}{2}$ into $(3)$,we get $\mu = 4(\frac{-3}{2}) + 1 = -6 + 1 = -5$.
Now,substitute $\lambda = \frac{-3}{2}$ and $\mu = -5$ into $(2)$:
$3(\frac{-3}{2}) - 2(-5) = k+1$
$\frac{-9}{2} + 10 = k+1$
$\frac{-9+20}{2} = k+1$
$\frac{11}{2} = k+1$
$k = \frac{11}{2} - 1 = \frac{9}{2}$.

(D) The equation of the line passing through $A(-2,-2,3)$ and parallel to the vector $\vec{v} = -2\hat{i} + 2\hat{j} - \hat{k}$ is given by:
$\frac{x+2}{-2} = \frac{y+2}{2} = \frac{z-3}{-1} = \lambda$
Any point on this line can be represented as $(-2\lambda - 2, 2\lambda - 2, -\lambda + 3)$.
Since the point $P$ lies on the $YOZ-$ plane,its $x-$coordinate must be $0$.
Setting the $x-$coordinate to $0$:
$-2\lambda - 2 = 0 \Rightarrow -2\lambda = 2 \Rightarrow \lambda = -1$.
Substituting $\lambda = -1$ into the general coordinates:
$x = -2(-1) - 2 = 0$
$y = 2(-1) - 2 = -4$
$z = -(-1) + 3 = 4$
Thus,the coordinates of point $P$ are $(0, -4, 4)$.

(C) The given Cartesian equations of the line are $y=2$ and $4x-3z+5=0$.
We can rewrite these as $y=2$ and $4x=3z-5$.
Dividing by $12$ to get the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$:
$4x = 3(z - \frac{5}{3}) \implies \frac{x}{3} = \frac{z - 5/3}{4}$.
Since $y=2$ is a constant,the direction ratio for $y$ is $0$.
Thus,the line can be written as $\frac{x-0}{3} = \frac{y-2}{0} = \frac{z-5/3}{4}$.
The line passes through the point $(0, 2, 5/3)$ and has direction ratios $(3, 0, 4)$.
The vector equation is $\overline{r} = \vec{a} + \lambda \vec{b}$,where $\vec{a} = 0\hat{i} + 2\hat{j} + \frac{5}{3}\hat{k}$ and $\vec{b} = 3\hat{i} + 0\hat{j} + 4\hat{k}$.
Therefore,$\overline{r} = (2\hat{j} + \frac{5}{3}\hat{k}) + \lambda(3\hat{i} + 4\hat{k})$.

(D) Given the Cartesian equation: $3x + 1 = 6y - 2 = -z + 1$.
First,rewrite the equation in the standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.
$3(x + \frac{1}{3}) = 6(y - \frac{1}{3}) = -(z - 1)$.
Dividing by the least common multiple of the coefficients $(6)$,we get:
$\frac{x + 1/3}{1/3} = \frac{y - 1/3}{1/6} = \frac{z - 1}{-1}$.
To simplify the direction ratios,multiply the denominators by $6$:
$\frac{x + 1/3}{2} = \frac{y - 1/3}{1} = \frac{z - 1}{-6}$.
The point on the line is $(-\frac{1}{3}, \frac{1}{3}, 1)$ and the direction vector is $2 \hat{i} + \hat{j} - 6 \hat{k}$.
Thus,the vector equation is $\overline{r} = (-\frac{1}{3} \hat{i} + \frac{1}{3} \hat{j} + \hat{k}) + \lambda(2 \hat{i} + \hat{j} - 6 \hat{k})$.

(D) Given Cartesian equations are $y=2$ and $4x-3z+5=0$.
From $4x-3z+5=0$,we have $4x = 3z-5$,which implies $4x = 3(z - \frac{5}{3})$.
This can be written as $\frac{x}{3} = \frac{z - 5/3}{4}$.
Since $y=2$,we can write the equation as $\frac{x-0}{3} = \frac{y-2}{0} = \frac{z-5/3}{4}$.
This line passes through the point $(0, 2, 5/3)$ and has direction ratios $(3, 0, 4)$.
The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is $\vec{r} = \vec{a} + \lambda \vec{b}$.
Here,$\vec{a} = 0\hat{i} + 2\hat{j} + \frac{5}{3}\hat{k}$ and $\vec{b} = 3\hat{i} + 0\hat{j} + 4\hat{k}$.
Thus,the vector equation is $\vec{r} = (2\hat{j} + \frac{5}{3}\hat{k}) + \lambda(3\hat{i} + 4\hat{k})$.

(A) The equation of the line passing through points $A(3, 5, -1)$ and $B(6, 3, -2)$ is given by the formula $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.
Substituting the coordinates,we get $\frac{x-3}{6-3} = \frac{y-5}{3-5} = \frac{z-(-1)}{-2-(-1)}$,which simplifies to $\frac{x-3}{3} = \frac{y-5}{-2} = \frac{z+1}{-1} = r$.
Any point $P$ on this line can be represented as $(3r+3, -2r+5, -r-1)$.
Given that the $y$-coordinate of point $P$ is $2$,we set $-2r+5 = 2$.
Solving for $r$,we get $-2r = -3$,so $r = \frac{3}{2}$.
Now,substitute $r = \frac{3}{2}$ into the $z$-coordinate expression: $z = -r-1 = -\frac{3}{2} - 1 = -\frac{5}{2}$.

(A) The equation of the line passing through $A(2, 4, 5)$ and $B(1, 2, 3)$ is given by:
$\frac{x-2}{1-2} = \frac{y-4}{2-4} = \frac{z-5}{3-5} = k$
$\frac{x-2}{-1} = \frac{y-4}{-2} = \frac{z-5}{-2} = k$
The coordinates of any point $P$ on this line are given by $(x, y, z) = (-k+2, -2k+4, -2k+5)$.
Given that the $z$-coordinate of point $P$ is $3$,we have:
$-2k + 5 = 3$
$-2k = -2$
$k = 1$
Substituting $k = 1$ into the expression for the $y$-coordinate:
$y = -2(1) + 4 = 2$.
Thus,the $y$-coordinate of point $P$ is $2$.

(D) The given lines are $\bar{r} = \bar{a}_1 + \lambda \bar{b}_1$ and $\bar{r} = \bar{a}_2 + \mu \bar{b}_2$.
Here,$\bar{a}_1 = \hat{i} + 2\hat{j} - \hat{k}$,$\bar{b}_1 = 2\hat{i} + \hat{j} - 3\hat{k}$,$\bar{a}_2 = 2\hat{i} - \hat{j} + 2\hat{k}$,and $\bar{b}_2 = \hat{i} - \hat{j} + \hat{k}$.
First,calculate $\bar{a}_2 - \bar{a}_1 = (2-1)\hat{i} + (-1-2)\hat{j} + (2-(-1))\hat{k} = \hat{i} - 3\hat{j} + 3\hat{k}$.
Next,calculate the cross product $\bar{b}_1 \times \bar{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1-3) - \hat{j}(2+3) + \hat{k}(-2-1) = -2\hat{i} - 5\hat{j} - 3\hat{k}$.
The magnitude is $|\bar{b}_1 \times \bar{b}_2| = \sqrt{(-2)^2 + (-5)^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$.
The shortest distance is $d = \left| \frac{(\bar{b}_1 \times \bar{b}_2) \cdot (\bar{a}_2 - \bar{a}_1)}{|\bar{b}_1 \times \bar{b}_2|} \right| = \left| \frac{(-2\hat{i} - 5\hat{j} - 3\hat{k}) \cdot (\hat{i} - 3\hat{j} + 3\hat{k})}{\sqrt{38}} \right| = \left| \frac{-2 + 15 - 9}{\sqrt{38}} \right| = \frac{4}{\sqrt{38}} = \frac{4}{\sqrt{2} \times \sqrt{19}} = \frac{2 \times 2}{\sqrt{2} \times \sqrt{19}} = \frac{2 \sqrt{2}}{\sqrt{19}}$ units.

(A) The given lines are $\overline{r}=\overline{a_1}+\lambda\overline{b_1}$ and $\overline{r}=\overline{a_2}+\mu\overline{b_2}$.
Here,$\overline{a_1}=2\hat{i}-\hat{j}$,$\overline{b_1}=2\hat{i}+\hat{j}-3\hat{k}$ and $\overline{a_2}=\hat{i}-\hat{j}+2\hat{k}$,$\overline{b_2}=2\hat{i}+\hat{j}-5\hat{k}$.
The vector $\overline{a_2}-\overline{a_1} = (\hat{i}-\hat{j}+2\hat{k})-(2\hat{i}-\hat{j}) = -\hat{i}+2\hat{k}$.
The cross product $\overline{b_1} \times \overline{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 2 & 1 & -5 \end{vmatrix} = \hat{i}(-5+3) - \hat{j}(-10+6) + \hat{k}(2-2) = -2\hat{i}+4\hat{j}$.
The magnitude $|\overline{b_1} \times \overline{b_2}| = \sqrt{(-2)^2+4^2+0^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$.
The shortest distance $d = \left| \frac{(\overline{a_2}-\overline{a_1}) \cdot (\overline{b_1} \times \overline{b_2})}{|\overline{b_1} \times \overline{b_2}|} \right|$.
$d = \left| \frac{(-\hat{i}+2\hat{k}) \cdot (-2\hat{i}+4\hat{j})}{2\sqrt{5}} \right| = \left| \frac{2+0+0}{2\sqrt{5}} \right| = \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}}$ units.

(D) The given lines are $L_1: \frac{x-3}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}$ and $L_2: \frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5}$.
The direction ratios of $L_1$ are $\vec{a_1} = (-3, 2k, 2)$.
The direction ratios of $L_2$ are $\vec{a_2} = (3k, 1, -5)$.
Since the lines are perpendicular,the dot product of their direction ratios must be zero:
$\vec{a_1} \cdot \vec{a_2} = 0$
$(-3)(3k) + (2k)(1) + (2)(-5) = 0$
$-9k + 2k - 10 = 0$
$-7k = 10$
$k = -\frac{10}{7}$.

(B) The direction ratios of the first line are $\vec{b_1} = (1, 2, 2)$.
The direction ratios of the second line are $\vec{b_2} = (3, 2, 6)$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19$.
Calculating the magnitudes: $|\vec{b_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$|\vec{b_2}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
Substituting these values into the formula: $\cos \theta = \frac{19}{3 \times 7} = \frac{19}{21}$.
Therefore,$\theta = \cos^{-1}\left(\frac{19}{21}\right)$.

(B) The equation of a line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Since the required line is parallel to the line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$,it will have the same direction ratios,which are $(3, 5, 6)$.
Given that the line passes through the point $(1, -3, 5)$,we substitute these values into the formula:
$\frac{x-1}{3} = \frac{y-(-3)}{5} = \frac{z-5}{6}$
This simplifies to $\frac{x-1}{3} = \frac{y+3}{5} = \frac{z-5}{6}$.
Thus,the correct option is $B$.

(B) The Cartesian equation of a line passing through a point $(x_1, y_1, z_1)$ and parallel to a vector $a \hat{i} + b \hat{j} + c \hat{k}$ is given by the formula:
$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$
Here,the point is $(x_1, y_1, z_1) = (5, -2, 4)$ and the direction vector is $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$,so $(a, b, c) = (3, 2, -8)$.
Substituting these values into the formula,we get:
$\frac{x - 5}{3} = \frac{y - (-2)}{2} = \frac{z - 4}{-8}$
Simplifying this,we obtain:
$\frac{x - 5}{3} = \frac{y + 2}{2} = \frac{z - 4}{-8}$
Thus,option $B$ is the correct answer.

(B) The first line is given by $\vec{r} = (3\hat{i} + \hat{j} - 2\hat{k}) + t(\hat{i} - \hat{j} - 2\hat{k})$. The direction vector is $\vec{b_1} = \hat{i} - \hat{j} - 2\hat{k}$ and it passes through $P_1(3, 1, -2)$.
The second line is given by $x = 4+k, y = -k, z = -4-2k$,which can be written as $\vec{r} = (4\hat{i} - 4\hat{k}) + k(\hat{i} - \hat{j} - 2\hat{k})$. The direction vector is $\vec{b_2} = \hat{i} - \hat{j} - 2\hat{k}$ and it passes through $P_2(4, 0, -4)$.
Since $\vec{b_1} = \vec{b_2}$,the lines are parallel.
To check if they are coincident,we check if $P_1$ lies on the second line. Substituting $P_1(3, 1, -2)$ into the second line equations:
$3 = 4+k \implies k = -1$
$1 = -k \implies k = -1$
$-2 = -4-2k \implies -2 = -4-2(-1) = -4+2 = -2$
Since $k = -1$ satisfies all equations,the point $P_1$ lies on the second line. Therefore,the lines are coincident.

(D) The direction ratios of the first line are $a_1 = 2, b_1 = 2, c_1 = 1$.
The direction ratios of the second line are $a_2 = 4, b_2 = 1, c_2 = 8$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$.
Substituting the values: $\cos \theta = \left| \frac{(2)(4) + (2)(1) + (1)(8)}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{4^2 + 1^2 + 8^2}} \right|$.
$\cos \theta = \left| \frac{8 + 2 + 8}{\sqrt{4 + 4 + 1} \sqrt{16 + 1 + 64}} \right| = \left| \frac{18}{\sqrt{9} \sqrt{81}} \right|$.
$\cos \theta = \frac{18}{3 \times 9} = \frac{18}{27} = \frac{2}{3}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{2}{3}\right)$.

(D) The Cartesian equation of a line passing through a point $(x_1, y_1, z_1)$ and parallel to a vector $\vec{b} = a \hat{i} + b \hat{j} + c \hat{k}$ is given by the formula:
$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$
Here,the point is $(x_1, y_1, z_1) = (5, 2, -4)$ and the direction vector is $\vec{b} = 3 \hat{i} + 2 \hat{j} - 8 \hat{k}$,so $a=3, b=2, c=-8$.
Substituting these values into the formula,we get:
$\frac{x-5}{3} = \frac{y-2}{2} = \frac{z-(-4)}{-8}$
$\frac{x-5}{3} = \frac{y-2}{2} = \frac{z+4}{-8}$
Thus,the correct option is $D$.

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

(C) The given lines are $\frac{x-1}{-3} = \frac{y-2}{1} = \frac{z-1}{2}$ and $\frac{x-2}{p} = \frac{y-1}{2} = \frac{z-2}{1}$.
The direction ratios of the first line are $\vec{a_1} = (-3, 1, 2)$.
The direction ratios of the second line are $\vec{a_2} = (p, 2, 1)$.
Since the lines are mutually perpendicular,the dot product of their direction ratios must be zero:
$\vec{a_1} \cdot \vec{a_2} = 0$
$(-3)(p) + (1)(2) + (2)(1) = 0$
$-3p + 2 + 2 = 0$
$-3p + 4 = 0$
$3p = 4$
$p = \frac{4}{3}$.

(D) The direction ratios of the first line are $\vec{b_1} = (1, 2, 2)$.
The direction ratios of the second line are $\vec{b_2} = (3, 2, 6)$.
The angle $\theta$ between two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ is given by $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.
Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19$.
Calculating the magnitudes: $|\vec{b_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$|\vec{b_2}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
Thus,$\cos \theta = \frac{19}{3 \times 7} = \frac{19}{21}$.
Therefore,$\theta = \cos^{-1}\left(\frac{19}{21}\right)$.

(B) The given Cartesian equation is $\frac{x-5}{3} = \frac{y+4}{7} = \frac{6-z}{2}$.
First,rewrite the equation in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
For the third term,$\frac{6-z}{2} = \frac{-(z-6)}{2} = \frac{z-6}{-2}$.
So,the equation becomes $\frac{x-5}{3} = \frac{y-(-4)}{7} = \frac{z-6}{-2}$.
Comparing this with the standard form,the point $(x_1, y_1, z_1)$ is $(5, -4, 6)$ and the direction ratios $(a, b, c)$ are $(3, 7, -2)$.
The position vector of the point is $\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k}$ and the direction vector is $\vec{b} = 3\hat{i} + 7\hat{j} - 2\hat{k}$.
The vector equation of a line is given by $\vec{r} = \vec{a} + \lambda\vec{b}$.
Substituting the values,we get $\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} - 2\hat{k})$.
Thus,the correct option is $B$.

(A) The direction ratios of the first line are $\vec{b_1} = (2, 3, 6)$.
The direction ratios of the second line are $\vec{b_2} = (10, 2, -11)$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Calculating the dot product: $a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(10) + (3)(2) + (6)(-11) = 20 + 6 - 66 = -40$.
Calculating the magnitudes: $|\vec{b_1}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
$|\vec{b_2}| = \sqrt{10^2 + 2^2 + (-11)^2} = \sqrt{100 + 4 + 121} = \sqrt{225} = 15$.
Thus,$\cos \theta = \frac{|-40|}{7 \times 15} = \frac{40}{105} = \frac{8}{21}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{8}{21}\right)$.

(D) The equation of a line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Given the point is $(2, 3, 4)$,so $x_1 = 2, y_1 = 3, z_1 = 4$.
The line is parallel to the $Y$-axis. The direction ratios of the $Y$-axis are $(0, 1, 0)$.
Therefore,the direction ratios of the line are $(a, b, c) = (0, 1, 0)$.
Substituting these values into the equation,we get $\frac{x-2}{0} = \frac{y-3}{1} = \frac{z-4}{0}$.
Thus,the correct option is $D$.

(D) The given lines are $\frac{x+3}{2}=\frac{y}{-3}=\frac{z+5}{-6}$ and $\frac{x-1}{10}=\frac{y+1}{-2}=\frac{z-3}{11}$.
The direction ratios of the first line are $\vec{b_1} = (2, -3, -6)$ and the direction ratios of the second line are $\vec{b_2} = (10, -2, 11)$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_1^2}}$.
Calculating the dot product: $a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(10) + (-3)(-2) + (-6)(11) = 20 + 6 - 66 = -40$.
Calculating the magnitudes: $|\vec{b_1}| = \sqrt{2^2 + (-3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
$|\vec{b_2}| = \sqrt{10^2 + (-2)^2 + 11^2} = \sqrt{100 + 4 + 121} = \sqrt{225} = 15$.
Thus,$\cos \theta = \frac{|-40|}{7 \times 15} = \frac{40}{105} = \frac{8}{21}$.
Therefore,$\theta = \cos^{-1}\left(\frac{8}{21}\right)$.

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

(C) The given equations of the lines are:
Line $1$: $\frac{2x-5}{k} = \frac{y+2}{-5} = \frac{z}{1}$.
Rewrite the first line in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$:
$\frac{2(x - 5/2)}{k} = \frac{y+2}{-5} = \frac{z}{1} \implies \frac{x - 5/2}{k/2} = \frac{y+2}{-5} = \frac{z}{1}$.
The direction ratios of the first line are $\vec{v_1} = (k/2, -5, 1)$.
Line $2$: $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$.
The direction ratios of the second line are $\vec{v_2} = (1, 2, 3)$.
Since the lines are perpendicular,the dot product of their direction ratios must be zero:
$(k/2)(1) + (-5)(2) + (1)(3) = 0$.
$k/2 - 10 + 3 = 0$.
$k/2 - 7 = 0$.
$k/2 = 7$.
$k = 14$.

(D) The equation of a line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Since the line passes through the origin,$(x_1, y_1, z_1) = (0, 0, 0)$.
Since the line is parallel to the $X$-axis,its direction ratios are proportional to $(1, 0, 0)$,so $(a, b, c) = (1, 0, 0)$.
Substituting these values into the equation,we get $\frac{x-0}{1} = \frac{y-0}{0} = \frac{z-0}{0}$,which simplifies to $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$.
Therefore,the correct option is $D$.