Let the line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the lines $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ at the points $A$ and $B$ respectively. Then the distance of the mid-point of the line segment $AB$ from the plane $2x-2y+z=14$ is

What is the reflection of the point $P(1, 3, 4)$ in the plane $2x - y + z + 3 = 0$?

Show that the points $(\hat{i}-\hat{j}+3 \hat{k})$ and $3(\hat{i}+\hat{j}+\hat{k})$ are equidistant from the plane $\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0$ and lie on opposite sides of it.

What is the length of the perpendicular drawn from the origin to the plane passing through point $a$ and containing the line $r = b + \lambda c$?

If $\lambda_1 < \lambda_2$ are two values of $\lambda$ such that the angle between the planes $P_1: \vec{r} \cdot (3 \hat{i} - 5 \hat{j} + \hat{k}) = 7$ and $P_2: \vec{r} \cdot (\lambda \hat{i} + \hat{j} - 3 \hat{k}) = 9$ is $\sin^{-1}\left(\frac{2 \sqrt{6}}{5}\right)$,then the square of the length of the perpendicular from the point $(38 \lambda_1, 10 \lambda_2, 2)$ to the plane $P_1$ is $...........$.

The plane containing the line $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and parallel to the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{4}$ passes through the point