Answer
$x=0$ or $x=-9$
Work Step by Step
2 by 2 determinant: $D=\left|\begin{array}{ll}
a & b\\
c & d
\end{array}\right|=ad-bc$
3 by 3 determinant: $D=\left|\begin{array}{lll}
{a_{11}}&{a_{12}}&{a_{13}}\\
{a_{21}}&{a_{22}}&{a_{23}}\\
{a_{31}}&{a_{32}}&{a_{33}}\end{array}\right|=$
$=a_{11}\left|\begin{array}{cc}
{a_{22}}&{a_{23}}\\
{a_{32}}&{a_{33}}\end{array}\right|-a_{12}\left|\begin{array}{cc}
{a_{21}}&{a_{23}}\\
{a_{31}}&{a_{33}}\end{array}\right|+a_{13}\left|\begin{array}{cc}
{a_{21}}&{a_{22}}\\{a_{31}}&{a_{32}}\end{array}\right|$
---
$\left|\begin{array}{rrr}{x}&{2}&{3}\\{1}&{x}&{0}\\{6}&{1}&{-2}\end{array}\right|=x\left|\begin{array}{rr}
{x}&{0}\\{1}&{-2}\end{array}\right|-2\left|\begin{array}{rr}{1}&{0}\\{6}&{-2}\end{array}\right|+3\left|\begin{array}{cc}{1}&{x}\\{6}&{1}\end{array}\right|$
$=x(-2x-0)-2(-2-0)+3(1-6x)$
$=-2x^{2}+4+3-18x$
$=-2x^{2}-18x+7$
So, we are solving
$-2x^{2}-18x+7=7$
$-2x^{2}-18x=0$
$-2x(x+9)=0$
$x=0$ or $x=-9$
