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Question
Given: Vectors:
- $\vec{a} = \lambda^2 \hat{i} + \hat{j} + \hat{k}$
- $\vec{b} = \hat{i} + \lambda^2 \hat{j} + \hat{k}$
- $\vec{c} = \hat{i} + \hat{j} + \lambda^2 \hat{k}$
Condition: Vectors are coplanar ⟹ Scalar triple product = 0
$\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$
Step 1: Use determinant:
$ \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} \lambda^2 & 1 & 1 \\ 1 & \lambda^2 & 1 \\ 1 & 1 & \lambda^2 \end{vmatrix} $
Step 2: Expand the determinant:
$ = \lambda^2(\lambda^2 \cdot \lambda^2 - 1 \cdot 1) - 1(1 \cdot \lambda^2 - 1 \cdot 1) + 1(1 \cdot 1 - \lambda^2 \cdot 1) \\ = \lambda^2(\lambda^4 - 1) - (\lambda^2 - 1) + (1 - \lambda^2) $
Simplify:
$= \lambda^6 - \lambda^2 - \lambda^2 + 1 + 1 - \lambda^2 = \lambda^6 - 3\lambda^2 + 2$
Step 3: Set scalar triple product to 0:
$\lambda^6 - 3\lambda^2 + 2 = 0$
Step 4: Let $x = \lambda^2$, then:
$x^3 - 3x + 2 = 0$
Factor:
$x^3 - 3x + 2 = (x - 1)^2(x + 2)$
So, $\lambda^2 = 1$ (double root), or $\lambda^2 = -2$ (discard as it's not real)
Thus, real values of $\lambda$ are: $\lambda = \pm1$
✅ Final Answer: $\boxed{2}$ distinct real values
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