The point collinear with $(1, -2, -3)$ and $(2, 0, 0)$ among the following is

$A$ line $L_1$ passes through the point with position vector $3 \hat{i}$ and is parallel to the vector $-\hat{i}+\hat{j}+\hat{k}$. Another line $L_2$ passes through the point with position vector $\hat{i}+\hat{j}$ and is parallel to the vector $\hat{i}+\hat{k}$. Find the position vector of the point of intersection of lines $L_1$ and $L_2$.

If the lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are perpendicular to each other,then $k = \dots$

Find the length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}$.

If $p$ is the shortest distance between the lines $\frac{x + 1}{7} = \frac{y + 1}{-6} = \frac{z + 1}{1}$ and $\vec{r} = (3\hat{i} + 5\hat{j} + 7\hat{k}) + \mu(\hat{i} - 2\hat{j} + \hat{k})$, then $[p]$ is... (where $[.]$ denotes the greatest integer function.)

$A$ line $L$ passes through points $A(1, 3, 2)$ and $B(2, 2, 1)$. If the mirror image of point $P(1, 1, -1)$ in the line $L$ is $(x, y, z)$,then $x+y+z=$