Given equation of circle is
$$
\begin{array}{r}
3 x^{2}+3 y^{2}-9 x+6 y+5=0 \\
\Rightarrow \quad x^{2}+y^{2}-3 x+2 y+\frac{5}{3}=0
\end{array}
Centre
=\left(\frac{3}{2},-1\right)
$$
and radius $=\sqrt{\frac{9}{4}+1-\frac{5}{3}}$
$$
=\sqrt{\frac{19}{12}}=\frac{1}{2} \sqrt{\frac{19}{3}}
$$
We know that, centre of the circle is the mid-point of the diameter.
Lives one and of point of diameter in (1,2) Let the other end point of diameter is $(h, k)$
Then, $\quad\left(\frac{3}{2},-1\right)=\left(\frac{1+h}{2}, \frac{2+k}{2}\right)$
$\Rightarrow \quad \frac{1+h}{2}=\frac{3}{2}$
$\Rightarrow 1+h=3$
$\Rightarrow h=2$
and $\frac{2+k}{2}=-1$
$\Rightarrow 2+k=-2$
$\Rightarrow k=-4$
So, the other end point is (2,-4)