If the line $\frac{x+1}{1}=\frac{y-k}{11}=\frac{z-4}{-5}$ lies in the plane $2x+py+7z-41=0$ which is perpendicular to the plane $x+4y-2z+13=0$,then $k=$

The foot of the perpendicular drawn from the point $(1, 3, 4)$ to the plane $2x - y + z + 3 = 0$ is:

Let the lines $L_{1}: \overrightarrow{r} = \lambda(\hat{i} + 2\hat{j} + 3\hat{k}), \lambda \in R$ and $L_{2}: \overrightarrow{r} = (\hat{i} + 3\hat{j} + \hat{k}) + \mu(\hat{i} + \hat{j} + 5\hat{k}), \mu \in R$ intersect at the point $S$. If a plane $ax + by - z + d = 0$ passes through $S$ and is parallel to both the lines $L_{1}$ and $L_{2}$,then the value of $a + b + d$ is equal to:

Let the lines $\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$ and $\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does $NOT$ lie on $P$?

The symmetric equation of the line formed by the intersection of the planes $3x + 2y + z - 5 = 0$ and $x + y - 2z - 3 = 0$ is:

The acute angle $\theta$ between the $xy$-plane and the plane passing through the point $(1, 2, 4)$ and parallel to the vectors with direction ratios $3, 2, -1$ and $1, -2, -2$ is