Answer
$x=0$ or, $x=-\dfrac{1}{2}$
Work Step by Step
The determinant for a 3 by 3 matrix can be computed as: $Determinant (D)= \begin{vmatrix} a & b & c \\
d & e & f \\
g & h & i \\ \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)$
We have:
$D= \begin{vmatrix}
x & 1 & 2 \\
1 & x & 3 \\
0 & 1 & 2 \\ \end{vmatrix}
$
Therefore, $D=x (2x-3)-1(2-0)+2(1-0) \\=2x^2-3x $
Since, $det=-4x$
So, $2x^2-3x=-4x \implies 2x^2-3x+4x=0$
$\implies x(2x+1)=0$
$\implies x=0$ or, $x=-\dfrac{1}{2}$
