The lines x = ay - 1 = z - 2 and x = 3y - 2 = bz - 2, (ab ≠ 0) are coplanar, if :

Solution

Maths sir · 4 steps

Use x itself as the parameter on both lines, and the coplanarity determinant collapses to one factor.

1. Read each line. On the first line put x = t. Then y = (t+1)/a and z = t + 2, so it passes through A(0, 1/a, 2) with direction (1, 1/a, 1). On the second line put x = s. Then y = (s+2)/3 and z = (s+2)/b, so it passes through B(0, 2/3, 2/b) with direction (1, 1/3, 1/b).

   |0, 2/3-1/a, 2/b-2; 1, 1/a, 1; 1, 1/3, 1/b| = 0

2. Expand the determinant. The first row is AB⃗, and the other two rows are the directions. Expanding along the first row gives -(2/3-1/a)(1/b-1) + 2(1/b-1)(1/3-1/a). Take out (1/b-1) and the bracket left over is -2/3+1/a+2/3-2/a = -1/a.
3. Solve the condition. The condition becomes -1/a(1/b-1) = 0. Since a ≠ 0, this forces b = 1, and a can be any non-zero value. As a check, with b = 1 both lines satisfy z - 2 = x, so both lie in the plane z = x + 2.
4. So. The lines are coplanar exactly when b = 1 and a ∈ ℝ - {0}, which is option (a).

The idea to keep. Two lines are coplanar when the vector joining their points and their two directions have a zero scalar triple product. Factor the determinant before solving and see which parameter actually survives.

Answer from the NTA final answer key. Solution by PrepWiser.