The equation of the straight line perpendicular to  5x-2y=7 and passing through point of intersection of the lines 2x+3y=6 and 3x+4y=6 is 

Point of intersection of

2x+3y-6=0 and 3x+4y=6 is

$\begin{array}{l} xy1\\ \begin{array}{cccc}3& -6& 2& 3\\ 4& -6& 3& 4\end{array}\end{array}$

$\begin{array}{l}\Rightarrow \frac{x}{-18+24}=\frac{y}{-18+12}=\frac{1}{8-9}\\ \Rightarrow \frac{x}{6}=\frac{y}{-6}=\frac{1}{-1}\\ \therefore \left(x,y\right)=\left(-6,6\right)\end{array}$

Let the line $\perp r$  to 5x-2y-7=0 is

        2x+5y+k=0---(1)

If  passes through (-6, 6)

$\begin{array}{l}\Rightarrow 2\left(-6\right)+5\left(6\right)+k=0\\ \Rightarrow -12+30+k=0\\ \Rightarrow k=-18\end{array}$

$\therefore$ Required equation is $2x+5y-18=0$