Equation of given lines
$$
f(x, y) \equiv 3 x^2-11 x y+10 y^2-7 x+13 y+4=0
$$

The point of intersection of line is same as the point of intersection of lines $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$


$$
\begin{array}{ll}
\therefore & \frac{\partial f}{\partial x}=6 x-11 y-7=0 \\
\text { and } & \frac{\partial f}{\partial y}=-11 x+20 y+13=0
\end{array}
$$
by cross multiplication method, we have
$$
\begin{aligned}
& \frac{x}{-143+140}=\frac{-y}{78-77}=\frac{1}{120-121} \\
\Rightarrow \quad & \frac{x}{3}=\frac{y}{1}=\frac{1}{1}
\end{aligned}
$$
$\therefore$ Point of intersection is $(3,1)$.