Given three sides of a triangle are
$$
\begin{array}{r}
x+8 y-22=0 ...(i)\\
5 x+2 y-34=0 ...(ii)\\
2 x-3 y+13=0 ...(iii)
\end{array}
$$
On solving Eqs. (i) and (ii), we get
$$
x_{1}=6, y_{1}=2
$$
On solving Eqs. (il) and (iii), we get
$$
x_{2}=4, y_{2}=7
$$
On solving Eqs. (i) and (iii), we get
$$
x_{3}=-2, y_{2}=3
$$
If the points of the vertices of the triangle are
$\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$. then the area of triangle $=\frac{1}{2}\left|\begin{array}{ccc}6 & 2 & 1 \\ 4 & 7 & 1 \\ -2 & 3 & 1\end{array}\right|$
$$
\begin{array}{l}
=\frac{1}{2}[6(7-3)-2(4+2)+1(12+14)] \\
=\frac{1}{2}(24-12+26)=\frac{1}{2}(12+26) \\
=\frac{1}{2} \times 38
\end{array}
$$
$=19 \mathrm{sq}$ units