If the line,frac{x-3}{2}=frac{y+2}{1}=frac{z+ | vedclass

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(D) Since the line lies in the plane,the direction vector of the line is perpendicular to the normal vector of the plane. The direction vector of the

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$\frac{122}{49}$

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(D) Since the line lies in the plane,the direction vector of the line is perpendicular to the normal vector of the plane. The direction vector of the line is $\vec{v} = 2\hat{i} + \hat{j} + 3\hat{k}$ and the normal to the plane is $\vec{n} = \ell\hat{i} + m\hat{j} - \hat{k}$.Thus,$\vec{v} \cdot \vec{n} = 0 \Rightarrow 2\ell + m - 3 = 0$,which gives $2\ell + m = 3$ (Equation $i$).Also,any point on the line must lie on the plane. The point $(3, -2, -4)$ lies on the line,so it must satisfy the plane equation $\ell x + my - z = 9$.Substituting the point: $3\ell - 2m - (-4) = 9 \Rightarrow 3\ell - 2m = 5$ (Equation $ii$).Solving the system of equations:From $(i)$,$m = 3 - 2\ell$. Substituting into $(ii)$:$3\ell - 2(3 - 2\ell) = 5 \Rightarrow 3\ell - 6 + 4\ell = 5 \Rightarrow 7\ell = 11 \Rightarrow \ell = \frac{11}{7}$.Then $m = 3 - 2(\frac{11}{7}) = \frac{21 - 22}{7} = -\frac{1}{7}$.Finally,$\ell^2 + m^2 = (\frac{11}{7})^2 + (-\frac{1}{7})^2 = \frac{121}{49} + \frac{1}{49} = \frac{122}{49}$.

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

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