Given equation,
$2 x^2+k x y-6 y^2+3 x+y+1=0$
is a pair of straight line.
$\therefore \quad\left|\begin{array}{ccc}2 & k / 2 & 3 / 2 \\ k / 2 & -6 & 1 / 2 \\ 3 / 2 & 1 / 2 & 1\end{array}\right|=0$
$\Rightarrow \quad\left|\begin{array}{ccc}4 & k & 3 \\ k & -12 & 1 \\ 3 & 1 & 2\end{array}\right|=0$
$=4(-24-1)-k(2 k-3)+3(k+36)=0$
$\Rightarrow \quad-100-2 k^2+3 k+3 k+108=0$
$\begin{array}{ll}\Rightarrow & 2 k^2-6 k-8=0 \\ \Rightarrow & k^2-3 k-4=0\end{array}$
$\Rightarrow \quad(k-4)(k+1)=0, \therefore k=4 \quad\{\because k>0\}$
$\therefore$ Equation of line is
$2 x^2+4 x y-6 y^2+3 x+y+1=0$
$\begin{array}{rrr}\Rightarrow & (2 x-2 y+1)(x+3 y+1) & =0 \\ \Rightarrow & 2 x-2 y+1=0 \text { and } x+3 y+1 & =0\end{array}$
Solving equation, we get
$x=\frac{-5}{8}, y=\frac{-1}{8}$
$\therefore$ Intersection point $\left(\frac{-5}{8}, \frac{-1}{8}\right)$.