Find the coordinates of the foot of the perpendicular drawn from the point $P(-1, 1, 2)$ to the plane $2x - 3y + z - 11 = 0$.

What is the distance between the line $\frac{x - 1}{3} = \frac{y + 2}{-2} = \frac{z - 1}{2}$ and the plane $2x + 2y - z = 6$?

The distance of the point $P(3, 8, 2)$ from the line $\frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3}$ measured parallel to the plane $3x + 2y - 2z + 15 = 0$ is (in $\text{ units}$)

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{b} = 2\hat{i} + 7\hat{j} + 3\hat{k}$. Let $L_1: \overrightarrow{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda \overrightarrow{a}, \lambda \in R$ and $L_2: \overrightarrow{r} = (\hat{j} + \hat{k}) + \mu \overrightarrow{b}, \mu \in R$ be two lines. If the line $L_3$ passes through the point of intersection of $L_1$ and $L_2$,and is parallel to $\overrightarrow{a} + \overrightarrow{b}$,then $L_3$ passes through the point:

Find the equation of the plane passing through the points $(2, 1, -1)$ and $(-1, 3, 4)$ and perpendicular to the plane $x - 2y + 4z = 10$.

The plane passing through the intersection of the planes $x + y + z = 1$ and $2x + 3y + z - 4 = 0$ and parallel to the $y$-axis also passes through the point: