Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z = 4$ and $2x + y - z = -5$,and perpendicular to the plane $5x + 3y + 6z + 8 = 0$.

On which of the following lines does the point of intersection of the line $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane $x+y+z=2$ lie?

The equation of a plane containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(1, 1, 0)$ is

The distance from the point $(-1, -5, -10)$ to the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 5$ is:

If the line joining the points $\hat{i}+\hat{j}$ and $3 \hat{i}+\hat{j}-\hat{k}$ meets the plane that passes through the point $2 \hat{i}+4 \hat{j}$ and is parallel to the vectors $3 \hat{j}+5 \hat{k}$ and $3 \hat{i}-\hat{k}$ at point $P$,then the position vector of the point $P$ is

Show that the lines $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ and $\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$ are coplanar.