If 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 mutually perpendicular,then $k$ is equal to

Let $L$ be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let $S$ be the set of all points $(a, b, c)$ on $L$, whose distance from the point $P(-1, -1, -9)$ is $7$. Then $\sum_{(a,b,c)\in S} (a+b+c)$ is equal to:

If the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{\lambda}$ and $\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-2}{3}$ are coplanar,then $\sin ^{-1}(\sin \lambda)+\cos ^{-1}(\cos \lambda)=$

Let $(\alpha, \beta, \gamma)$ be the coordinates of the foot of the perpendicular drawn from the point $(5, 4, 2)$ on the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$. Then the length of the projection of the vector $\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ on the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$ is:

Find the coordinates of the foot of the perpendicular from the point $(2, 4, 1)$ to the line $\vec{r} = (-5, -3, 6) + k(1, 4, -9)$,where $k \in R$.

The angle between the lines $\frac{x-2}{2} = \frac{y-3}{-2} = \frac{z-5}{1}$ and $\frac{x-2}{1} = \frac{y-3}{2} = \frac{z-5}{2}$ is $ . . . . . . $. (in $^{\circ}$)