Step-by-step Solution:

The vectors \( \mathbf{v}_1 = \mathbf{i} + \lambda \mathbf{j} + \mathbf{k} \), \( \mathbf{v}_2 = \lambda \mathbf{i} + \mathbf{j} + \mathbf{k} \), and \( \mathbf{v}_3 = 2\mathbf{i} + \lambda \mathbf{j} + \mathbf{k} \) are coplanar if their scalar triple product is zero.

The scalar triple product is given by:

\[ \mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3) = 0 \]

First, compute \( \mathbf{v}_2 \times \mathbf{v}_3 \):

\[ \mathbf{v}_2 \times \mathbf{v}_3 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \lambda & 1 & 1 \\ 2 & \lambda & 1 \end{vmatrix} \]

Expanding the determinant:

\[ \mathbf{v}_2 \times \mathbf{v}_3 = \mathbf{i} \begin{vmatrix} 1 & 1 \\ \lambda & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} \lambda & 1 \\ 2 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} \lambda & 1 \\ 2 & \lambda \end{vmatrix} \]

Calculate the minors:

\[ \begin{vmatrix} 1 & 1 \\ \lambda & 1 \end{vmatrix} = (1)(1) - (\lambda)(1) = 1 - \lambda \]

\[ \begin{vmatrix} \lambda & 1 \\ 2 & 1 \end{vmatrix} = (\lambda)(1) - (2)(1) = \lambda - 2 \]

\[ \begin{vmatrix} \lambda & 1 \\ 2 & \lambda \end{vmatrix} = (\lambda)(\lambda) - (2)(1) = \lambda^2 - 2 \]

Thus:

\[ \mathbf{v}_2 \times \mathbf{v}_3 = (1 - \lambda)\mathbf{i} - (\lambda - 2)\mathbf{j} + (\lambda^2 - 2)\mathbf{k} \]

Next, compute \( \mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3) \):

\[ \mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3) = (1)(1 - \lambda) + (\lambda)(-(\lambda - 2)) + (1)(\lambda^2 - 2) \]

Simplify the expression:

\[ \mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3) = 1 - \lambda - \lambda^2 + 2\lambda + \lambda^2 - 2 \]

\[ \mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3) = 1 - \lambda + 2\lambda - 2 = \lambda - 1 \]

For the vectors to be coplanar:

\[ \lambda - 1 = 0 \]

Thus, \( \lambda = 1 \).

There is only 1 distinct value of \( \lambda \) for which the vectors are coplanar.