Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3},$ we get:

$\left|\begin{array}{ccc}3 x+a & 3 x+a & 3 x+a \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$

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

Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1},$ we have:

$\Rightarrow(3 x+a)\left|\begin{array}{lll}1 & 1 & 1 \\ x & a & x \\ x & x & a\end{array}\right|=0$

Expanding along $R_{1},$ we have:

$(3 x+a)\left[1 \mathrm{x} a^{2}\right]=0$

$\Rightarrow a^{2}(3 x+a)=0$

But $a \neq 0$

Therefore, we have:

$3 x+a=0$

$\Rightarrow x=-\frac{a}{3}$