Question

The equation of the plane passing through the point \((1,2,1)\) and perpendicular to the planes \(x+2y+2z-7 = 0\) and \(3x+3y+2z-5 = 0\) is

• A. \(2x+y-2z-2 = 0\)
• B. \(2x-4y+3z+3 = 0\)
• C. \(2x+4y-5z-5 = 0\)
• D. \(x+4y-6z-3 = 0\)

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 B

Step 1: Conditions
Normal $(a, b, c)$ satisfies $a + 2b + 2c = 0$ and $3a + 3b + 2c = 0$.

Step 2: Solve
Subtract: $2a + b = 0$, so $b = -2a$. Then $a - 4a + 2c = 0$ gives $c = 3a/2$. Take $a = 2$: normal $(2, -4, 3)$.

Step 3: Through the point
$2(1) - 4(2) + 3(1) + d = 0$ gives $d = 3$. Plane: $2x - 4y + 3z + 3 = 0$, option (B).

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