The interval contained in the domain of definition of non-zero solutions of the differential equation $(x-3)^2 y'+y=0$, is

$(-\pi / 2, \pi / 2)$

We have,

$(x-3)^2 y'+y=0$

$\Rightarrow (x-3)^2 \frac{d y}{d x}+y=0 \Rightarrow \frac{1}{y} d y+\frac{1}{(x-3)^2} d x=0$

Integrating, we get

$\log |y|-\frac{1}{x-3}=\log C$

$\Rightarrow \log \left\{\frac{|y|}{C}\right\}=\frac{1}{x-3} \Rightarrow|y|=C e^{\frac{1}{x-3}}$

The domain of definition of this solution is $R-\{3\}$.

Clearly, $(-\pi / 2, \pi / 2) \subset R-\{3\}$. So, option (a) is correct.