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

Line is perpendicular to normal to plane
\(\begin{aligned}
& \Rightarrow(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(l \hat{\mathrm{i}}+\mathrm{m} \hat{\mathrm{j}}-\hat{\mathrm{k}})=0 \\
& 2 l+\mathrm{m}-3=0...(i)
\end{aligned}\)
\((3,-2,-4)\) lies on the plane
\(l x+\mathrm{m} y-\mathrm{z}=9\)
\(\begin{aligned}
& \therefore \quad 3 l-2 \mathrm{~m}+4=9 \\
& \Rightarrow 3 l-2 \mathrm{~m}=5...(ii)
\end{aligned}\)
Solving (i) and (ii), we get
\(\begin{aligned}
& l=\frac{11}{7}, \mathrm{~m}=\frac{-1}{7} \\
& l^2+\mathrm{m}^2\left(\frac{11}{7}\right)^2+\left(\frac{-1}{7}\right)^2=\frac{122}{49}
\end{aligned}\)