<p>The shortest distance from the plane \(12x + 4y + 3z = 327\) to the sphere \(x^2 + y^2 + z^2 + 4x - 2y - 6z = 155\) is</p>
Step-by-Step Solution
Key Concept: Find the sphere's center and radius, calculate the perpendicular distance from center to plane, then subtract the radius to get the shortest distance from plane to sphere surface.
Step 1: Convert sphere to standard form x^2 + y^2 + z^2 + 4x - 2y - 6z = 155 (x^2 + 4x + 4) + (y^2 - 2y + 1) + (z^2 - 6z + 9) = 155 + 4 + 1 + 9 (x + 2)^2 + (y - 1)^2 + (z - 3)^2 = 169 Center C = (-2, 1, 3), Radius r = 13 Step 2: Find perpendicular distance from center to plane Plane: 12x + 4y + 3z = 327 Distance d = |12(-2) + 4(1) + 3(3) - 327| / √(12^2 + 4^2 + 3^2) d = |-24 + 4 + 9 - 327| / √(144 + 16 + 9) d = |-338| / √169 d = 338 / 13 = 26 Step 3: Calculate shortest distance from plane to sphere Shortest distance = d - r = 26 - 13 = 13 ∴ Answer: C
Correct Answer: C