The number of values of $k$ for which the linear equations

4x + ky + 2z = 0

kx + 4y + z = 0

2x + 2y + z = 0

posses a non-zero solution is

A homogeneous system has a non-trivial solution $\iff$ the determinant of its coefficient matrix is $0$.
Coefficient matrix $A=\begin{bmatrix}4 & k & 2\\ k & 4 & 1\\ 2 & 2 & 1\end{bmatrix}$. Hence, $$ \det(A)= \begin{vmatrix} 4 & k & 2\\ k & 4 & 1\\ 2 & 2 & 1 \end{vmatrix} =-(k-4)(k-2). $$ Setting $\det(A)=0 \Rightarrow -(k-4)(k-2)=0 \Rightarrow k=2 \text{ or } k=4.$
Therefore, the number of values of $k$ is $\boxed{2}$.