The equations of perpendicular bisectors of sides AB,BC of  triangle ABC are  $\mathrm{x}-3\mathrm{y}-5=0,\mathrm{x}+2\mathrm{y}=0$ respectively and A(1,2)  then C is 

The vertex  B(h,k) is the image of  A(1,2) with respect to  $\mathrm{x}-3\mathrm{y}-5=0$
Use the formula 
$\begin{array}{l}\frac{h-1}{1}=\frac{k-2}{-3}=-\frac{2(1-6-5)}{1^2+3^2}\left(\because \frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-2\left(a{x}_1+b{y}_1+c\right)}{a^2+b^2}\right)\\ \frac{h-1}{1}=\frac{k-2}{-3}=\frac{20}{10}\\ \frac{h-1}{1}=\frac{k-2}{-3}=2\end{array}$            
It implies 
$\begin{array}{l}\frac{\mathrm{h}-1}{1}=2\\ \mathrm{h}-1=2\\ \mathrm{h}=3\end{array}$ and $\begin{array}{l}\frac{\mathrm{k}-2}{-3}=2\\ \mathrm{k}-2=-6\\ \mathrm{k}=-4\end{array}$
Therefore, B(3,-4) 
The image of  B(3,-4) with respect to the line $\mathrm{x}+2\mathrm{y}=0$ is the point  C
Use the formula 
$\begin{array}{l}\frac{\mathrm{h}-3}{1}=\frac{\mathrm{k}+4}{2}=-\frac{2(3-8)}{1^2+2^2}\bullet \frac{\mathrm{h}-{\mathrm{x}}_1}{\mathrm{a}}=\frac{\mathrm{k}-{\mathrm{y}}_1}{\mathrm{b}}=\frac{-2\left({\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c}\right)}{{\mathrm{a}}^2+{\mathrm{b}}^2}\\ \frac{\mathrm{h}-3}{1}=\frac{\mathrm{k}+4}{2}=2\\ \frac{\mathrm{h}-3}{1}=\frac{\mathrm{k}+4}{2}=2\end{array}$       
It implies 
$\begin{array}{l}\frac{\mathrm{h}-3}{1}=2\\ \mathrm{h}-3=2\\ \mathrm{h}=5\end{array}$
and 
 $\begin{array}{l}\frac{\mathrm{k}+4}{2}=2\\ \mathrm{k}+4=4\\ \mathrm{k}=0\end{array}$ 
Therefore, C(5,0)