The points $A(3, 2, 0)$,$B(5, 3, 2)$,and $C(-9, 6, -3)$ are the vertices of a triangle $ABC$. If $AD$ is the internal bisector of $\angle BAC$ which meets $BC$ at $D$,then the coordinates of $D$ are:

If the image of the point $P(1, -2, 1)$ with respect to the line passing through the points $B(1, 1, 2)$ and $C(2, 2, 1)$ is $R(l, m, n)$, then $l^2 + m^2 + n^2 =$

If $L_1$ is a line through the point $5 \hat{i}+8 \hat{j}+11 \hat{k}$ and parallel to the vector $2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $L_2$ is a line through the point $4 \hat{i}+6 \hat{j}+8 \hat{k}$ and parallel to the vector $3 \hat{i}+4 \hat{j}+5 \hat{k}$, then the point of intersection of $L_1$ and $L_2$ is

If the lines $\frac{x-k}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\frac{9}{2}}{2}=\frac{z}{1}$ intersect,then the value of $k$ is

Let $Q$ be the cube with the set of vertices $\{(x_1, x_2, x_3) \in \mathbb{R}^3: x_1, x_2, x_3 \in \{0,1\}\}$. Let $F$ be the set of all twelve lines containing the diagonals of the six faces of the cube $Q$. Let $S$ be the set of all four lines containing the main diagonals of the cube $Q$; for instance,the line passing through the vertices $(0,0,0)$ and $(1,1,1)$ is in $S$. For lines $\ell_1$ and $\ell_2$,let $d(\ell_1, \ell_2)$ denote the shortest distance between them. Then the maximum value of $d(\ell_1, \ell_2)$,as $\ell_1$ varies over $F$ and $\ell_2$ varies over $S$,is

The Cartesian equation of a line is $3x + 1 = 6y - 2 = -z + 1$. Find its vector equation.