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The line x/1 = y/2 = z/3 intersects the pair of planes given by 3x + 3(1-2*alpha)*y + z = 3 and 6*alpha*x + 3(1-2*beta)*y + 2z = -2 (derived from the combined equation). Find the plane on which the point (alpha, beta, 1) lies.

1. 2x - y + z = 4
2. x + y - z = 2
3. x - 2y = 0
4. 2x - y = 0

Correct answer: 2x - y = 0

Solution

Substituting x=t, y=2t, z=3t into the two plane equations gives two equations in t, alpha and beta. The intersection condition yields specific values of alpha and beta which, when checked against the four option planes at (alpha, beta, 1), identify the correct plane.

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