d
$|A-x I|=0 \Rightarrow\left|\begin{array}{cc}1-x & 5 \\ \lambda & 10-x\end{array}\right|=0$
$\Rightarrow x^2-11 x+10-5 \lambda=0$
$\Rightarrow(10-5 \lambda) A^{-1}=-A+11 I$
$\alpha+\beta=-2 \Rightarrow \frac{10}{10-5 \lambda}=-2 \Rightarrow 10-5 \lambda=-5 \Rightarrow \lambda=3$
$\therefore \alpha=\frac{1}{5} \quad \quad \beta=\frac{-11}{5}$
$\therefore 4 a^2+\beta^2+\lambda^2=\frac{4}{25}+\frac{121}{25}+3^2=14 \text { Ans. }$