A matrix is singular if its determinant is zero. The determinant of a 3×3 matrix can be calculated using the formula below:
$$\det \left[ {\begin{array}{{20}{c}} a&b&c \ d&e&f \ g&h&i \end{array}} \right] = (-1)^{a+b+c} \left| \begin{array}{{20}{c}} a&c \ e&i \end{array} \right| + b \left| \begin{array}{{20}{c}} b&f \ d&h \end{array} \right| + c \left| \begin{array}{{20}{c}} c&d \ e&g \end{array} \right|$$
In this case, the matrix is:
$$\left[ {\begin{array}{*{20}{c}} 8&{\text{x}}&0 \ 4&0&2 \ {12}&6&0 \end{array}} \right]$$
Therefore, the determinant is:
$$\det \left[ {\begin{array}{{20}{c}} 8&{\text{x}}&0 \ 4&0&2 \ {12}&6&0 \end{array}} \right] = (-1)^{8+0+0} \left| \begin{array}{{20}{c}} 8&0 \ 12&0 \end{array} \right| + 0 \left| \begin{array}{{20}{c}} 8&2 \ 12&6 \end{array} \right| + 0 \left| \begin{array}{{20}{c}} 8&2 \ 4&6 \end{array} \right| = 0$$
The determinant is zero when $x=0$. Therefore, the matrix becomes singular when $x=0$.