Step-by-step Solution:

1. Condition for Coplanarity: Three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar if and only if their scalar triple product is zero. The scalar triple product, denoted as $[\vec{a} \ \vec{b} \ \vec{c}]$, is given by the determinant of the matrix formed by their components.

   The given vectors are: $\vec{a} = - {\lambda ^2}\widehat i + \widehat j + \widehat k$ $\vec{b} = \widehat i - {\lambda ^2}\widehat j + \widehat k$ $\vec{c} = \widehat i + \widehat j - {\lambda ^2}\widehat k$

   The condition for these vectors to be coplanar is: $\begin{vmatrix} -\lambda^2 & 1 & 1 \\ 1 & -\lambda^2 & 1 \\ 1 & 1 & -\lambda^2 \end{vmatrix} = 0$

2. Evaluating the Determinant: To simplify the determinant, we can perform column operations. Let's apply the operation $C_1 \to C_1 + C_2 + C_3$: $\begin{vmatrix} -\lambda^2 + 1 + 1 & 1 & 1 \\ 1 - \lambda^2 + 1 & -\lambda^2 & 1 \\ 1 + 1 - \lambda^2 & 1 & -\lambda^2 \end{vmatrix} = 0$ $\begin{vmatrix} 2 - \lambda^2 & 1 & 1 \\ 2 - \lambda^2 & -\lambda^2 & 1 \\ 2 - \lambda^2 & 1 & -\lambda^2 \end{vmatrix} = 0$ Now, we can factor out the common term $(2 - \lambda^2)$ from the first column: $(2 - \lambda^2) \begin{vmatrix} 1 & 1 & 1 \\ 1 & -\lambda^2 & 1 \\ 1 & 1 & -\lambda^2 \end{vmatrix} = 0$ Next, we simplify the remaining determinant using row operations. Let's apply $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: $(2 - \lambda^2) \begin{vmatrix} 1 & 1 & 1 \\ 1-1 & -\lambda^2-1 & 1-1 \\ 1-1 & 1-1 & -\lambda^2-1 \end{vmatrix} = 0$ $(2 - \lambda^2) \begin{vmatrix} 1 & 1 & 1 \\ 0 & -\lambda^2-1 & 0 \\ 0 & 0 & -\lambda^2-1 \end{vmatrix} = 0$ The determinant of an upper or lower triangular matrix is the product of its diagonal elements. So, we expand along the first column: $(2 - \lambda^2) [1 \cdot ((- \lambda^2 - 1)(- \lambda^2 - 1) - 0 \cdot 0)] = 0$ $(2 - \lambda^2) (-\lambda^2 - 1)^2 = 0$ $(2 - \lambda^2) (-1)^2 (\lambda^2 + 1)^2 = 0$ $(2 - \lambda^2) (\lambda^2 + 1)^2 = 0$

3. Solving for $\lambda$: The equation gives two possibilities:

   Case 1: $2 - \lambda^2 = 0$ $\lambda^2 = 2$ This gives two distinct real values for $\lambda$: $\lambda = \sqrt{2}$ and $\lambda = -\sqrt{2}$.

   Case 2: $(\lambda^2 + 1)^2 = 0$ $\lambda^2 + 1 = 0$ $\lambda^2 = -1$ This equation has no real solutions for $\lambda$, as the square of a real number cannot be negative. The solutions are $\lambda = \pm i$, which are complex.

4. Counting the Distinct Real Values: The question asks for the number of distinct real values of $\lambda$. From our analysis, the only real values are $\sqrt{2}$ and $-\sqrt{2}$. Therefore, there are two distinct real values of $\lambda$.

5. Conclusion: The number of distinct real values of $\lambda$ for which the given vectors are coplanar is 2. This corresponds to option C.