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

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(B) The given line is $\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z+4}{3}$.The direction vector of the line is $\vec{v} = (2, -1, 3)$ and a point on the l

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(B) The given line is $\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z+4}{3}$.The direction vector of the line is $\vec{v} = (2, -1, 3)$ and a point on the line is $P(3, -2, -4)$.The equation of the plane is $lx + my - z = 9$,which has a normal vector $\vec{n} = (l, m, -1)$.Since the line lies in the plane,the normal vector $\vec{n}$ must be perpendicular to the direction vector $\vec{v}$ of the line. Thus,$\vec{n} \cdot \vec{v} = 0$:$2l - m - 3 = 0 \Rightarrow 2l - m = 3$  ....$(1)$Also,the point $P(3, -2, -4)$ must satisfy the equation of the plane:$l(3) + m(-2) - (-4) = 9$$3l - 2m + 4 = 9$$3l - 2m = 5$  ....$(2)$Solving equations $(1)$ and $(2)$:From $(1)$,$m = 2l - 3$. Substituting into $(2)$:$3l - 2(2l - 3) = 5$$3l - 4l + 6 = 5$$-l = -1 \Rightarrow l = 1$Substituting $l = 1$ into $(1)$:$m = 2(1) - 3 = -1$Therefore,$l^2 + m^2 = (1)^2 + (-1)^2 = 1 + 1 = 2$.

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

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