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 \((2,3,4)\) is

- A plane perpendicular to two given planes passes through \((1,2,1)\). The distance of the plane from the point \((2,3,4)\).
1. Let the plane equation passing through \((1,2,1)\) be:
\(a(x-1)+b(y-2)+c(z-1)=0\)
2. The plane is perpendicular to the two given planes:
- \(2 x-2 y+z=0\)
- \(x-y+2 z=4\)
Thus, the normal vector of the required plane is perpendicular to the normal vectors of these planes.
3. Solve the normal vectors using the cross-product method:
- Normal to plane 1: \(\overrightarrow{n_1}=(2,-2,1)\)
- Normal to plane 2: \(\overrightarrow{n_2}=(1,-1,2)\)
- Cross product \(\overrightarrow{n_1} \times \overrightarrow{n_2}\) gives the required normal.
4. Use the distance formula for the point \((2,3,4)\) to calculate the final distance from the plane.
5. Answer: Option \(2 \frac{2 \sqrt{2}}{5}\) units.