Number of Real Values of Lambda for Coplanar Vectors

To find the number of distinct real values of $\lambda$ for which the given vectors are coplanar, we set their scalar triple product equal to zero.

Let the three vectors be:
$\vec{a} = -\lambda^{2} \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} - \lambda^{2} \hat{j} + \hat{k}$
$\vec{c} = \hat{i} + \hat{j} - \lambda^{2} \hat{k}$

The condition for these vectors to be coplanar is:

$$[\vec{a} \quad \vec{b} \quad \vec{c}] = \begin{vmatrix} -\lambda^{2} & 1 & 1 \\ 1 & -\lambda^{2} & 1 \\ 1 & 1 & -\lambda^{2} \end{vmatrix} = 0$$

Let $x = -\lambda^{2}$. The determinant becomes:

$$\Delta = \begin{vmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}$$

Applying the row operation $R_1 \to R_1 + R_2 + R_3$:

$$\Delta = \begin{vmatrix} x+2 & x+2 & x+2 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix} = (x+2) \begin{vmatrix} 1 & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}$$

Applying the column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:

$$\Delta = (x+2) \begin{vmatrix} 1 & 0 & 0 \\ 1 & x-1 & 0 \\ 1 & 0 & x-1 \end{vmatrix}$$

Expanding along the first row:
$\Delta = (x+2)(x-1)^2$

For coplanarity, we set $\Delta = 0$:
$(x+2)(x-1)^2 = 0 \implies x = -2 \quad \text{or} \quad x = 1$

Now, substituting back $x = -\lambda^{2}$:

1. Case 1:
   $-\lambda^{2} = -2 \implies \lambda^{2} = 2 \implies \lambda = \pm\sqrt{2}$
   Both $\sqrt{2}$ and $-\sqrt{2}$ are real numbers.

2. Case 2:
   $-\lambda^{2} = 1 \implies \lambda^{2} = -1 \implies \lambda = \pm i$
   These values are non-real complex numbers.

Thus, the distinct real values of $\lambda$ are $\sqrt{2}$ and $-\sqrt{2}$.

Therefore, the number of distinct real values of $\lambda$ is two.

Correct Option: C