Reduced row echelon with imaginary numbers Working on the following problem:
Let $v = \begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix}
w = \begin{bmatrix} -3 \\ i \\ 8 \\ \end{bmatrix}
y = \begin{bmatrix} h \\ -5i \\ -3 \\ \end{bmatrix}$
For what values of $h$ is the vector $y$ in the plane generated by the vectors $v$ and $w$?
My work so far: I know that for $y$ to be in the plane generated by $v$ and $w$, $y$ has to be a linear combination of $v$ and $w$. This means that I need to find the solution of the augmented matrix:
$$
    \begin{bmatrix}
    1 & -3 & h \\
    0 & i & -5i \\
    -2 & 8 & -3 \\
    \end{bmatrix}
$$
I assume that I need to find the reduced echelon form of this matrix. However, I don't know how to deal with the imaginary numbers when doing the row reduction. Could anyone give me some pointers? 

If  the vector $y$ lies in the plane generated by the vectors $v$ and $w$ then $y$ can be expressed as a linear combination of $v$ and $w$ i.e. $y=c_1v+c_2w$.
Then we must have:
$$
    \begin{vmatrix}
    1 & -3 & h \\
    0 & i & -5i \\
    -2 & 8 & -3 \\
    \end{vmatrix}=0
$$ which gives the value of $h$.
