3 x 4 Matrix with Complex Numbers I have a $3 \times4$ matrix  $$
         \begin{pmatrix}
         i & 4c & -1 & 0 \\
         2ic & 4 & -2c & 0 \\
         -i & -2 & 1 & 6c-3 \\
         \end{pmatrix}
 $$
to figure out the complex numbers $c$ such that the row space has dimension 2, how would I approach this problem? Also, for each $c$ I would like to try to find the subset of the columns of A that makes it a $\Bbb C$-basis of the column-space A.
So far, i've taken the determinate's of the 3 sub $3 \times3$ matrices. In particular:
$$(A_1) = \begin{pmatrix}
         4c & -1 & 0 \\
         4 & -2c & 0 \\
         -2 & 1 & 6c-3 \\
         \end{pmatrix} = det(A_1) -48c^2+48c-12$$ 
$$(A_2) = \begin{pmatrix}
         i & -1 & 0 \\
         2ic & -2 & 0 \\
         -2 & 1 & 6c-3 \\
         \end{pmatrix} = det(A_2) 12ic^2-18ic-12i$$ 
$$(A_2) = \begin{pmatrix}
         i & 4c & -1 \\
         2ic & 4 & -2 \\
         -i & -2 & 1 \\
         \end{pmatrix} = det(A_3) -8c^2i+12ic-4i$$ 
From here, I'm not sure what to do.I set the determinate equal to zero and got the following (below). 
$$c(A_1)=\frac{1}{2}$$ $$c(A_2)= \frac{3i \pm 5i}{4i}$$ $$c(A_3)= \frac{-3i\pm i}{4}$$
Does anyone have any clue how or where I have messed up because I feel as though  I have made a mistake somewhere.

I think reducing by rows can help:$${}$$
$$ \begin{pmatrix}
         i & 4c & -1 & 0 \\
         2ic & 4 & -2c & 0 \\
         -i & -2 & 1 & 6c-3 \\
         \end{pmatrix}\longrightarrow\begin{pmatrix}
         i & 4c &\!\!-1 & 0 \\
         0 & 4-8c^2 & 0 & 0 \\
         0 & 4c-2 & 0 & 6c-3 \\
         \end{pmatrix}$$
Observe that if $\;6c-3=0\iff c=\frac12\;$ , then it follows that the last row becomes all zeros and none of the two first one is all zeros,  and also if $\;8c^2=4\iff c=\pm\frac1{\sqrt2}\;$, and with this value the second row is all zeros.
Thus, only for $\;c=\frac12,\,\pm\frac1{\sqrt2}\;$ the matrix has row dimension equal to two
