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

$5x-2y=7$  perpendicular line equation is

$-2x-5y+k=0\to \left(1\right)$

$2x+5y-k=0$

$2x+3y=6\to \left(2\right)\times 3$

$3x+4y=6\to \left(3\right)\times 2$

(2) × 3 – (3) × 2

$6x+9y=18$

$6x+8y=12$

$\underset{\_}{---}$

           y = 6

$2x+3y=6$

$2x=6-18$

$x=\frac{-12}{2}=-6$

$\left(x,y\right)=\left(-6,6\right)$

$2x+5y-k=0$  passes through (-6,6)

$2\left(-6\right)+5\left(6\right)-k=0$

$-12+30=k$

$18=k$

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