Question

If A and B are the feet of the perpendiculars drawn from \((1,2,3)\) to planes \(YZ\) and \(ZX\), then the equation of the plane passing through the points A, B and the origin is

• A. \(6x+3y+2z = 0\)
• B. \(6x-3y-2z = 0\)
• C. \(6x+3y-2z = 0\)
• D. \(3x+6y+2z = 0\)

Hint:

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

Answer 1

The Correct Option is C

Step 1: Test the options:
The normal of the plane must have zero dot product with $(1, 2, 2)$ and $(3, 3, 2)$.

Step 2: Check (B):
Normal $(2, -4, 3)$: with $(1, 2, 2)$: $2 - 8 + 6 = 0$. With $(3, 3, 2)$: $6 - 12 + 6 = 0$. And the point: $2 - 8 + 3 + 3 = 0$. All conditions hold. The other options fail, for example (A) has normal $(2, 1, -2)$ with dot product $2 + 2 - 4 = 0$ for the first plane but $6 + 3 - 4 = 5$ for the second.

Final Answer:
Option (B). \[ \boxed{2x - 4y + 3z + 3 = 0} \]