Answer
$det\left(A\right)=0$
Work Step by Step
$Theorem\:\left(1\right).\:Let\:A\:be\:a\:square\:matrix\:of\:order\:n.\:Then\:the\:determinant\:of\:A\:is\:given\:by\\det\:\left(A\right)=\sum _{j=1}^na_{ij}C_{ij}=a_{i1}C_{i1}+a_{i2}C_{i2}+...+a_{in}C_{in}\:\left(ith\:row\:expansion\right)\\Or\:det\:\left(A\right)=\sum _{i=1}^na_{ij}C_{ij}=a_{1j}C_{1j}+a_{2j}C_{2j}+...+a_{nj}C_{nj}\:\left(ith\:column\:expansion\right)\\ $
$\\In\:first\:column\:two\:elements\:are\:zero,\:so\:first\:column\:expansion\:to\:find\:determinant\\det\left(A\right)=\left(5\right)C_{11}+\left(4\right)C_{21}+\left(0\right)C_{31}+\left(0\right)C_{41}\\C_{11}=\left(-1\right)^{1+1}\begin{vmatrix}6&4&12\:\\ \:2&-3&4\:\\ \:1&-2&2\end{vmatrix}=\begin{vmatrix}6&4&12\:\\ \:2&-3&4\:\\ \:1&-2&2\end{vmatrix}\\Expanding\:by\:cofactors\:in\:the\:first\:row\:yields\\C_{11}=\left(6\right)\left(-1\right)^{1+1}\begin{vmatrix}-3&4\:\\ \:-2&2\end{vmatrix}+\left(4\right)\left(-1\right)^{1+2}\begin{vmatrix}2&4\:\\ \:1&2\end{vmatrix}+\left(12\right)\left(-1\right)^{^{1+3}}\begin{vmatrix}2&-3\\ 1&-2\end{vmatrix}=6\begin{vmatrix}-3&4\\ \:-2&2\end{vmatrix}-4\begin{vmatrix}2&4\\ \:1&2\end{vmatrix}+12\begin{vmatrix}2&-3\\ \:1&-2\end{vmatrix}\\=\left(6\right)\left[\left(-3\right)\left(2\right)-\left(4\right)\left(-2\right)\right]-4\left[\left(2\right)\left(2\right)-\left(4\right)\left(1\right)\right]+12\left[\left(2\right)\left(-2\right)-\left(-3\right)\left(1\right)\right]\\=\left(6\right)\left(2\right)-\left(4\right)\left(0\right)+12\left(-1\right)\\C_{11}=0\\$
$\\The\:cofactor\:of\:element\:second\:row\:first\:column\:element\:is\\C_{21}=\left(-1\right)^{2+1}\begin{vmatrix}3&0&6\\ 2&-3&4\\ 1&-2&2\end{vmatrix}=-\begin{vmatrix}3&0&6\\ 2&-3&4\\ 1&-2&2\end{vmatrix}\\In\:first\:row\:has\:single\:zero\:entry\\Expanding\:by\:cofactors\:in\:the\:first\:row\:yields\\C_{21}=\left(3\right)\left(-1\right)^{1+1}\begin{vmatrix}-3&4\\ \:-2&2\end{vmatrix}-\left(0\right)\left(-1\right)^{1+2}\begin{vmatrix}2&4\\ \:1&2\end{vmatrix}+6\left(-1\right)^{1+3}\begin{vmatrix}2&-3\\ \:1&2\end{vmatrix}\\=\left(3\right)\begin{vmatrix}-3&4\\ \:-2&2\end{vmatrix}-\left(0\right)\begin{vmatrix}2&4\\ \:1&2\end{vmatrix}+6\begin{vmatrix}2&-3\\ \:1&2\end{vmatrix}\\=\left(3\right)\left[\left(-3\right)\left(2\right)-\left(4\right)\left(-2\right)\right]-0+6\left[\left(2\right)\left(-2\right)-\left(-3\right)\left(1\right)\right]\\=\left(3\right)\left(2\right)-\left(4\right)\left(0\right)+6\left(-1\right)\\C_{21}=0\\det\left(A\right)=\left(5\right)C_{11}+\left(4\right)C_{21}+\left(0\right)C_{31}+\left(0\right)C_{41}\\det\left(A\right)=\left(5\right)\left(0\right)+\left(4\right)\left(0\right)+\left(0\right)C_{31}+\left(0\right)C_{41}\\det\left(A\right)=0$
