If the system of equations ${\left(a+2\right)}^{3}x+{\left(a+7\right)}^{3}y+{\left(a+15\right)}^{3}z=0,$ $\left(a+2\right)x+\left(a+7\right)y+\left(a+15\right)z=0$ and $x+y+z=0$ possesses a non-trivial solution, then $a=$_____

For non-trivial solution $\Delta =0$$\Rightarrow$$\left|\begin{array}{ccc}{\left(a+2\right)}^3& {\left(a+7\right)}^3& {\left(a+15\right)}^3\\ a+2& a+7& a+15\\ 1& 1& 1\end{array}\right|=0$

$$\begin{gathered} We know that \left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right) \\ By u\sin g this formula \\ \left(\left(a+2\right)-\left(a+7\right)\right)\left(\left(a+7\right)-\left(a+15\right)\right)\left(a+15-\left(a+2\right)\right)\left(a+2+a+7+a+15\right)=0 \\ \Rightarrow 3a+24=0 \\ \Rightarrow 3a=-24 \\ \Rightarrow a=-8 \end{gathered}$$

$3a+24=0\Rightarrow a=-8$