A plane which is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, passes through $(1,-2,1)$. The distance of the plane from the point $(1,2,2)$ is

Let $\mathrm{L}_1$ $\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+1}{1}$ and $\mathrm{L}_2: \frac{x-2}{2}=\frac{y+2}{1}=\frac{z-3}{3}$ be the given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

The equation of the plane passing through the point $(1,1,1)$ and perpendicular to the planes $2 x-y-2 z=5$ and $3 x-6 y+2 z=7$ is

Equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is