Answer:
Option C
Explanation:
Let $\alpha,\beta ,\gamma$ are roots of equation
$x^{3}-ax^{2}+ax-1$=0 ..........(i)
$\therefore$ $\alpha$+$\beta$+$\gamma$=a
$\alpha\beta +\beta\gamma+\alpha\gamma=a$
$\alpha\beta \gamma=-1$
Cubic equation whose roots $\alpha^{2}$, $\beta^{2}$, $\gamma ^{2}$ is
$x^{2}-(\alpha^{2}+\beta^{2}+ \gamma^{2})x^{2}+(\alpha^{2} \beta^{2}+\beta^{2}\gamma^{2}+\alpha^{2} \gamma^{2})x-\alpha ^{2}\beta^{2} \gamma ^{2}=0$ ........(ii)
Equi .(i) and (ii) are identical.
$\therefore\frac{a}{\alpha^{2}+\beta^{2}+ \gamma^{2}}=\frac{a}{\alpha^{2}\beta^{2}+\beta^{2}\gamma^{2}+ \alpha^{2}\gamma^{2}}=\frac{1}{\alpha^{2}\beta^{2} \gamma^{2}}$
a= $ \alpha ^{2}+\beta ^{2}+\gamma ^{2}$ [ $\alpha \beta \gamma=-1$]
a= $ (\alpha +\beta +\gamma )^{2}$-$2(\alpha \beta + \beta \gamma +\gamma \alpha )$
a= $ a^{2}-2a \Rightarrow a^{2}=3a$
$\Rightarrow$ a=3 [ $\because$ a is non-zero real ]
