Given $A=\left[\begin{array}{ccc}2& 0& -\alpha \\ 5& \alpha & 0\\ 0& \alpha & 3\end{array}\right]$ for $\alpha \in R-\{a,b\}anda>b A^{-1}$ exists and $A^{-1}=A^2-5bA+cI$ where $\alpha =1.$ The value of $a+5b+c$ is

$A=\left[\begin{array}{ccc}2& 0& -\alpha \\ 5& \alpha & 0\\ 0& \alpha & 3\end{array}\right];\left|A\right|=\left|\begin{array}{ccc}2& 0& -\alpha \\ 5& \alpha & 0\\ 0& \alpha & 3\end{array}\right|\ne 0$

$6\alpha -5{\alpha }^2\ne 0\Rightarrow \alpha (6-5\alpha )\ne 0$

$\alpha =0,\frac{6}{5}$

$\therefore \alpha \in R-\{0,\frac{6}{5}\}$

For $\alpha =1\Rightarrow A=\left[\begin{array}{ccc}2& 0& -1\\ 5& 1& 0\\ 0& 1& 3\end{array}\right]\Rightarrow \left|A\right|=6-5=1;AdjA=\left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right]$

$\therefore A^{-1}=\left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right]$

By characteristic equation $|A-xI|=0$

$\left|\begin{array}{ccc}2-x& 0& -1\\ 5& 1-x& 0\\ 0& 1& 3-x\end{array}\right|=0\Rightarrow x^3-6x^2+11x-1=0$ By cayley-hamilton theorem we get 

$A^3-6A^2+11A-I\Rightarrow A^{-1}=A^2-6A+11.I$