Let the third vertex of $\triangle A B C$ be $(a, b)$.
Orthocentre $=H(0,0)$
Let $A(5,-1)$ and $B(-2,3)$ be other two vertices of $\triangle A B C$.
Now, (Slope of $A H) \times($ Slope of $B C)=-1$
$$
\begin{aligned}
& \Rightarrow\left(\frac{-1-0}{5-0}\right)\left(\frac{b-3}{a+2}\right)=-1 \\
& \Rightarrow b-3=5(a+2)
\end{aligned}
$$
Similarly,
$($ Slope of $B H) \times($ Slope of $A C)=-1$
$$
\begin{aligned}
& \Rightarrow \quad-\left(\frac{3}{2}\right) \times\left(\frac{b+1}{a-5}\right)=-1 \\
& \Rightarrow \quad 3 b+3=2 a-10 \\
& \Rightarrow 3 b-2 a+13=0
\end{aligned}
$$
On solving equation (1) and (2) we get $a=-4, b=-7$
Hence, third vertex is $(-4,-7)$.