<p>For lines <br/>
\(\frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{\lambda^2}\)<br/>
and<br/>
\(\frac{x-3}{1} = \frac{y-2}{\lambda^2} = \frac{z-1}{2}\)<br/>
to be coplanar, the number of values of \(\lambda\) is:</p>
Step-by-Step Solution
Key Concept: Two lines are coplanar if and only if the scalar triple product of (point difference vector) with both direction vectors equals zero. This condition translates to a determinant equation in λ that you must solve.
Step 1: Write the direction vectors and a point on each line. Line 1: Point P_1 = (1, 2, -3), Direction d_1 = (1, 2, λ^2) Line 2: Point P_2 = (3, 2, 1), Direction d_2 = (1, λ^2, 2) Step 2: For coplanarity, the vectors P_1P_2, d_1, and d_2 must be coplanar. P_1P_2 = (2, 0, 4) The condition is: [P_1P_2 · (d_1 × d_2)] = 0, which means the determinant: $\begin{vmatrix} 2 & 0 & 4 \\ 1 & 2 & \lambda^2 \\ 1 & \lambda^2 & 2 \end{vmatrix} = 0$ Step 3: Expand along the first row: $2\begin{vmatrix} 2 & \lambda^2 \\ \lambda^2 & 2 \end{vmatrix} - 0 + 4\begin{vmatrix} 1 & 2 \\ 1 & \lambda^2 \end{vmatrix} = 0$ $2(4 - \lambda^4) + 4(\lambda^2 - 2) = 0$ $8 - 2\lambda^4 + 4\lambda^2 - 8 = 0$ $-2\lambda^4 + 4\lambda^2 = 0$ $-2\lambda^2(\lambda^2 - 2) = 0$ Step 4: Solve for λ: $\lambda^2 = 0 \text{ or } \lambda^2 = 2$ $\lambda = 0, \sqrt{2}, -\sqrt{2}$ ∴ Answer: C (The number of values of λ is 3 )
Correct Answer: C