<p>If the equation of the plane passing through the point \((-1, 2, 0)\) and parallel to the lines \(\dfrac{x}{3} = \dfrac{y+1}{0} = \dfrac{z-2}{-1}\) and \(\dfrac{x-1}{1} = \dfrac{y+1}{2} = \dfrac{z+1}{-1}\) is \(ax + by + cz = 1\), then the value of \((a + b + c)\) is:</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 point to determine the plane equation.
Step 1: Identify direction vectors of the two lines. Line 1: d_1 = (3, 0, -1) Line 2: d_2 = (1, 2, -1) Step 2: Find the normal vector to the plane using cross product n = d_1 × d_2 . n = | i j k | = i (0 + 2) - j (-3 + 1) + k (6 - 0) |3 0 -1| |1 2 -1| n = (2, 2, 6) or simplified (1, 1, 3) Step 3: Plane equation using normal (1, 1, 3) and point (-1, 2, 0): 1(x + 1) + 1(y - 2) + 3(z - 0) = 0 x + y + 3z - 1 = 0 x + y + 3z = 1 Step 4: Compare with ax + by + cz = 1: a = 1, b = 1, c = 3 ∴ a + b + c = 5
Correct Answer: C