The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is:

The distance of the point whose position vector is $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $r \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=4$ is

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 equation of the plane passing through the intersection of the planes $2x - 5y + z = 3$ and $x + y + 4z = 5$ and parallel to the plane $x + 3y + 6z = 1$ is $x + 3y + 6z = k$,where $k$ is:

If the line $\frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z + 4}{3}$ lies in the plane $lx + my - z = 9$,then $l^2 + m^2 = \dots$

Find the distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$.