<p>Let the line \(\dfrac{x-2}{3} = \dfrac{y-1}{-5} = \dfrac{z+2}{2}\) lie in the plane \(x + 3y - \alpha z + \beta = 0\). Then \((\alpha, \beta)\) equals</p>
Step-by-Step Solution
Key Concept: For a line to lie in a plane, two conditions must hold: (1) the direction vector of the line must be perpendicular to the normal vector of the plane, and (2) any point on the line must satisfy the plane equation.
Step 1: Identify the direction vector of the line and normal vector of the plane. Direction vector of line: d = (3, -5, 2) Normal vector of plane: n = (1, 3, -α) Step 2: Apply perpendicularity condition (d · n = 0). 3(1) + (-5)(3) + 2(-α) = 0 3 - 15 - 2α = 0 -12 - 2α = 0 α = -6 Step 3: Take a point on the line and substitute into the plane equation. When the parameter equals 0: Point P = (2, 1, -2) Substituting into x + 3y - αz + β = 0: 2 + 3(1) - (-6)(-2) + β = 0 2 + 3 - 12 + β = 0 -7 + β = 0 β = 7 ∴ Answer: (α, β) = (-6, 7)
Correct Answer: A