Question

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

• A. 1
• B. $\sqrt{2}$
• C. $2\sqrt{2}$
• D. $\sqrt{3}$

Hint:

The normal of the required plane is the cross product of the normals of the two given planes.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
A plane perpendicular to two planes has a normal along the cross product of their normals.

Step 2: Normal
\((2,-2,1)\times(1,-1,2)=(-4+1,\ -(4-1),\ -2+2)=(-3,-3,0)\), so the normal is along \((1,1,0)\).

Step 3: Equation
Through \((1,-2,1)\): \(1(x-1)+1(y+2)=0\), which is \(x+y+1=0\).

Step 4: Distance
Distance of \((1,2,2)\): \(\frac{|1+2+1|}{\sqrt{1^2+1^2}}=\frac{4}{\sqrt2}=2\sqrt2\).

Step 5: Compare the options
The distance is \(2\sqrt{2}\) units, the third option.

Final Answer:
The distance is \(2\sqrt{2}\) units. \[ \boxed{\text{(C) }2\sqrt{2}} \]