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.

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Cross product of normals gives required plane normal.
Updated On: Apr 26, 2026
  • 1
  • $\sqrt{2}$
  • $2\sqrt{2}$
  • $\sqrt{3}$
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The Correct Option is B

Solution and Explanation

Concept:
Plane perpendicular to two planes → normal is cross product of normals. Step 1: Normals. \[ (2,-2,1), (1,-1,2) \]
Step 2: Cross product. \[ = (-3,-3,0) \]
Step 3: Equation of plane. \[ -3(x-1)-3(y+2)=0 \]
Step 4: Distance. \[ d = \sqrt{2} \] Conclusion. Distance = $\sqrt{2}$
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