<p>The plane passing through the point (4, –1, 2) and parallel to the lines \(\dfrac{x+2}{3} = \dfrac{y-2}{-1} = \dfrac{z+1}{2}\) and \(\dfrac{x-2}{1} = \dfrac{y-3}{2} = \dfrac{z-4}{3}\) also passes through the point:</p>
Step-by-Step Solution
Key Concept: A plane parallel to two lines must have its normal vector perpendicular to both direction vectors; find the normal using the cross product of the two direction vectors, then use the given point to write the plane equation.
Step 1: Identify direction vectors of the two lines. Line 1: d_1 = (3, -1, 2) Line 2: d_2 = (1, 2, 3) Step 2: Find the normal vector to the plane using cross product n = d_1 × d_2 . n = | i j k | |3 -1 2| |1 2 3| n = i (-3 - 4) - j (9 - 2) + k (6 + 1) n = (-7, -7, 7) or simplified: (-1, -1, 1) Step 3: Write the plane equation using point (4, -1, 2) and normal (-1, -1, 1). -1(x - 4) - 1(y + 1) + 1(z - 2) = 0 -x + 4 - y - 1 + z - 2 = 0 -x - y + z + 1 = 0 or x + y - z - 1 = 0 Step 4: Check which point satisfies this equation. Substitute options into x + y - z - 1 = 0 to find the answer point. ∴ Answer: D
Correct Answer: D