The equation of the line passing through the point of intersection of the lines x+3y-1=0, x-2y+4=0 and perpendicular to the line 3x+2y=0 is

Given lines x+3y-1=0----(i)

   x-2y+4=0----(ii)

 3x+2y=0----(iii)

Point of intersection of (i) and (ii)

$\begin{array}{l}x y 1\\ \begin{array}{cccc}3& -1& 1& 3\\ -2& 4& 1& -2\end{array}\end{array}$

$\begin{array}{l}\Rightarrow \frac{x}{12-2}=\frac{y}{-1-4}=\frac{1}{-2-3}\\ \Rightarrow \frac{x}{10}=\frac{y}{-5}=\frac{1}{-5}\\ \therefore x=\frac{10}{-5}=-2,y=\frac{-5}{-5}=1\end{array}$

$\therefore$The point is (-2, 1)

Equation of a line $\perp r to 3x+2y=0 is 2x-3y+k=0$

It passes through  (-2, 1) $\Rightarrow 2\left(-2\right)-3\left(1\right)+k=0$

$\Rightarrow k=7$

$\therefore$equation is $2x-3y+7=0$