If the line, $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane $lx+my-z=9,$ then $l^2+m^2$ is equal to

$\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in $lx+my-z=9$

The point $\left(3,-2,-4\right)$ lies on the plane. So it should satisfy the equation of the plane.

$3l-2m+4=9$

$3l-2m=5$ .......(i)

$\therefore$ Line lies in plane

$\therefore$ line is perpendicular to the normal to the plane

$2\left(l\right)-1.\left(m\right)-3=0$

$2l-m=3$ .......(ii)

$l=1$ and $m= -1$

$\therefore$ $l^2+m^2=1+1=2$