Answer
$0,1,2$
Work Step by Step
Step 1. Start from the original, expand the determinant using the first column:
$\begin{array}( \\LHS= \\ \\ \end{array} \begin{vmatrix} x&12&13\\0&x-1&23\\0&0&x-2 \end{vmatrix}=x\begin{vmatrix} x-1&23\\0&x-2 \end{vmatrix}$
Step 2. Complete the expansion to get: $LHS=x(x-1)(x-2)=0$
Step 3. The solutions to the above equation can be found as: $x=0,1,2$
