<p>The distance of the point \((1, -2, 4)\) from the plane passing through the point \((1, 2, 2)\) and perpendicular to the planes \(x - y + 2z = 3\) and \(2x - 2y + z + 12 = 0\) is</p>
Step-by-Step Solution
Key Concept: The required plane is perpendicular to two given planes, so its normal vector is parallel to the cross product of the normals of the two given planes. Once you find the plane equation, use the point-to-plane distance formula.
Step 1: Find the normal vector of the required plane. Normal vectors of given planes are n_1 = (1, -1, 2) and n_2 = (2, -2, 1). Step 2: The normal to the required plane is n = n_1 × n_2 : n = | i j k | |1 -1 2| |2 -2 1| = i (-1 + 4) - j (1 - 4) + k (-2 + 2) = 3 i + 3 j + 0 k = (3, 3, 0) Step 3: Simplify: n = (1, 1, 0). The plane equation is 1(x - 1) + 1(y - 2) + 0(z - 2) = 0 ⟹ x + y - 3 = 0 Step 4: Distance from (1, -2, 4) to plane x + y - 3 = 0: d = |1(1) + 1(-2) - 3|/√(1^2 + 1^2) = |1 - 2 - 3|/√2 = |-4|/√2 = 4/√2 = 2√2 ∴ Answer: B
Correct Answer: B