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

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Aicharya · Accepted Answer

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

If the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane $\ell x+\mathrm{m}…

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