Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $5y + 8 = 0$.

The equation of the plane passing through the intersection of the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-5}{-3}$ and $\frac{x+5}{3}=\frac{y-4}{-1}=\frac{z+3}{4}$ and parallel to the $xy$-plane is

The line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is parallel to the plane

The distance of the point $P(1, 0, -3)$ from the plane $x - y - z = 9$ measured parallel to the line $\frac{x - 2}{2} = \frac{y + 2}{3} = \frac{z - 6}{-6}$ is (in $\text{ units}$)

Find the equation of the plane passing through the line of intersection of the planes $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$ and parallel to the $x$-axis.

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is . . . . . . units.