The length of the perpendicular from the point $(1, -2, 5)$ to the line passing through $(1, 2, 4)$ and parallel to the line $x + y - z = 0 = x - 2y + 3z - 5$ is:

If the lines $\frac{x-k}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+3}{1}$ are co-planar,then the value of $k$ is $.....$

$ABC$ is a triangle in a plane with vertices $A(2, 3, 5)$,$B(-1, 3, 2)$,and $C(\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the coordinate axes,then the value of $\lambda + \mu$ is:

Find the shortest distance between the lines given by $\vec{r}=(8+3 \lambda) \hat{i}+(-9-16 \lambda) \hat{j}+(10+7 \lambda) \hat{k}$ and $\vec{r}=15 \hat{i}+29 \hat{j}+5 \hat{k}+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k})$. (in $\text{ units}$)

If the distance of point $B(2, 1, -3)$ from the line passing through the point $A(4, -2, 2)$ and parallel to the vector $\vec{c} = -4\hat{i} - 6\hat{j} - 2\hat{k}$ is $x$, then $x^4 + x^2 + 541 =$

Find the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$.