The number of values of $k$ for which the linear equations $4x+ky+2z=0$, $kx+4y+z=0$, and $2x+2y+z=0$ possess a non-zero solution is:

For a system of homogeneous linear equations to have a non-zero solution, the determinant of the coefficient matrix must be zero ($D=0$).
$$D=\left|\begin{array}{ccc}4& k& 2\\ k& 4& 1\\ 2& 2& 1\end{array}\right|=0$$
Expanding along the first row:
$$4(4-2)-k(k-2)+2(2k-8)=0$$
$$4(2)-k^2+2k+4k-16=0$$
$$8-k^2+6k-16=0$$
$$-k^2+6k-8=0$$
$$k^2-6k+8=0$$
$$(k-4)(k-2)=0$$
The values of $k$ are $4$ and $2$. Thus, there are $2$ such values.