If$x\ne \frac{3}{2}$  and$\left|\begin{array}{ccc}2x-5& 1& 1\\ 1& 2x-5& 1\\ 1& 1& 2x-5\end{array}\right|=0$  then x =

$C_1\to C_1+C_2+C_3$

$\left|\begin{array}{ccc}2x-3& 1& 1\\ 2x-3& 2x-5& 1\\ 2x-3& 1& 2x-5\end{array}\right|=0$

 $\Rightarrow \left(2x-3\right) \left|\begin{array}{ccc}1& 1& 1\\ 1& 2x-5& 1\\ 1& 1& 2x-5\end{array}\right|=0$                              

2x-3=0 (or) $1[{(2x-5)}^2-1]-1[2x-5-1]+1\left[1-2x+5\right]=0$

$x=\frac{3}{2}$  (or) $4x^2-20x+24-2x+6+6-2x=0$             

Since $x\ne \frac{3}{2}$ $4x^2-24x+36=0$ $\Rightarrow$ $x^2-6x+9=0$ $\Rightarrow$  x = 3