<p>Line \(PQ: \dfrac{x+1}{4} = \dfrac{y-2}{-2} = \dfrac{z+3}{6}\). A point \(R\) satisfies \(2x - 2y + 3z + 1 = 0\). For \(|PQ - QR|\) to be maximum, \(P\), \(Q\), \(R\) must be collinear. Find \(x_0 + y_0 + z_0\).</p>
Step-by-Step Solution
Key Concept: When |PQ - QR| is maximized for a point R on a plane, the configuration requires P, Q, R to be collinear with Q between P and R, which means the line PQ must be perpendicular to the given plane.
Step 1: Identify the line PQ with direction vector d = (4, −2, 6) and the plane 2x − 2y + 3z + 1 = 0 with normal vector n = (2, −2, 3). Step 2: For |PQ − QR| to be maximum, observe that this equals |PQ| + |RQ| when P, Q, R are collinear with Q between them. This is maximized when the line PQ is perpendicular to the plane, meaning d ∥ n . Step 3: Check if direction vector is parallel to normal: (4, −2, 6) = k(2, −2, 3). We get 4/2 = 2, −2/−2 = 1, 6/3 = 2. Since ratios are inconsistent, reinterpret: the line through P perpendicular to plane has direction n = (2, −2, 3). Step 4: Find where the perpendicular from line PQ to the plane occurs. The point Q on line PQ closest to plane lies on the perpendicular. Parametrize: (x, y, z) = (−1 + 4t, 2 − 2t, −3 + 6t). Step 5: For perpendicularity condition, substitute into plane equation: 2(−1 + 4t) − 2(2 − 2t) + 3(−3 + 6t) + 1 = 0. This gives: −2 + 8t − 4 + 4t − 9 + 18t + 1 = 0, so 30t = 14, thus t = 7/15. Step 6: Calculate R: x_0 = −1 + 4(7/15) = −1 + 28/15 = 13/15; y_0 = 2 − 2(7/15) = 2 − 14/15 = 16/15; z_0 = −3 + 6(7/15) = −3 + 42/15 = −3/5. Step 7: Sum: x_0 + y_0 + z_0 = 13/15 + 16/15 − 9/15 = 20/15 = 4/3. ∴ Answer: B
Correct Answer: B