<p>A plane containing the point \((3, 2, 0)\) and the line \(\dfrac{x-1}{1} = \dfrac{y-2}{5} = \dfrac{z-3}{4}\) also contains the point</p>
Step-by-Step Solution
Key Concept: A plane containing a given point and a line must satisfy two conditions: the point lies on the plane, and every point on the line (including its direction vector) must satisfy the plane equation. Use the given point and line to determine the plane, then check which option satisfies it.
Step 1: Identify given information. The plane passes through point P(3, 2, 0) and contains the line with point A(1, 2, 3) and direction vector d = (1, 5, 4). Step 2: Find the normal vector to the plane. The normal n must be perpendicular to both the direction vector d = (1, 5, 4) and vector AP = (3-1, 2-2, 0-3) = (2, 0, -3). Step 3: Calculate n = d × AP = (1, 5, 4) × (2, 0, -3) = (-15, 11, -10). Step 4: Write the plane equation using point P(3, 2, 0) and normal (-15, 11, -10): -15(x-3) + 11(y-2) - 10(z-0) = 0 → -15x + 11y - 10z + 23 = 0. Step 5: Verify that point A(1, 2, 3) satisfies: -15(1) + 11(2) - 10(3) + 23 = -15 + 22 - 30 + 23 = 0 ✓ Step 6: Test each option by substituting into -15x + 11y - 10z + 23 = 0 to find which point lies on the plane. ∴ Answer: D
Correct Answer: D