Given line $2 x+3 y=k$
$$
\Rightarrow \frac{2 x+3 y}{k}=1 ...(i)
$$
Also given, $3 x^2-x y+3 y^2+2 x-3 y-4=0$
Now homogenising the above equation
$$
\begin{aligned}
& \Rightarrow 3 x^2-x y+3 y^2+2 x(1)-3 y(1)-4(1)^2=0 \\
& \Rightarrow 3 x^2-x y+3 y^2+2 x\left(\frac{2 x+3 y}{k}\right)-3 y\left(\frac{2 x+3 y}{k}\right) \\
& -4\left(\frac{2 x+3 y}{k}\right)^2=0 \\
& \Rightarrow 3 x^2-x y+3 y^2+\frac{\left(4 x^2+6 x y\right)}{k}-\frac{\left(6 x y+9 y^2\right)}{k} \\
& -\frac{4\left(4 x^2+9 y^2+12 x y\right)}{k^2}=0 \\
& 3 x^2 k^2-x y k^2+3 y^2 k^2+4 x^2 k+6 x y k-6 x y k \\
& \Rightarrow \frac{-9 y^2 k-16 x^2-36 y^2-48 x y}{k^2}=0 \\
&
\end{aligned}
$$
$$
\Rightarrow x^2\left(3 k^2+4 k-16\right)+y^2\left(3 k^2-9 k-36\right)+x y\left(-k^2-48\right)=0
$$

The above equation represents the pair of equation, According to question, these lines perpendicular to each other
Therefore,
$\left(\right.$ Coefficient of $\left.x^2\right)+\left(\right.$ Coefficient of $\left.y^2\right)=0$
$$
\begin{aligned}
& \Rightarrow\left(3 k^2+4 k-16\right)+\left(3 k^2-9 k-36\right)=0 \\
& \Rightarrow 6 k^2-5 k-52=0
\end{aligned}
$$