Let P be a plane lx + my + nz = 0 containing | vedclass

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(C) The line is given by $\frac{x-1}{-1} = \frac{y+4}{2} = \frac{z+2}{3}$.Since the plane $lx + my + nz = 0$ contains this line,it passes through the

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$2$

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(C) The line is given by $\frac{x-1}{-1} = \frac{y+4}{2} = \frac{z+2}{3}$.Since the plane $lx + my + nz = 0$ contains this line,it passes through the point $(1, -4, -2)$ and its normal vector $(l, m, n)$ is perpendicular to the direction vector $(-1, 2, 3)$ of the line.Thus,$l(1) + m(-4) + n(-2) = 0 \implies l - 4m - 2n = 0$ (Equation $1$).And $-l + 2m + 3n = 0$ (Equation $2$).Adding $(1)$ and $(2)$,we get $-2m + n = 0 \implies n = 2m$.Substituting $n = 2m$ into $(2)$,we get $-l + 2m + 3(2m) = 0 \implies l = 8m$.Thus,the plane equation is $8x + y + 2z = 0$.The point $C$ dividing $AB$ in ratio $k:1$ is $\left(\frac{2k-3}{k+1}, \frac{4k-6}{k+1}, \frac{-3k+1}{k+1}\right)$.Since $C$ lies on the plane,$8\left(\frac{2k-3}{k+1}\right) + \left(\frac{4k-6}{k+1}\right) + 2\left(\frac{-3k+1}{k+1}\right) = 0$.$16k - 24 + 4k - 6 - 6k + 2 = 0$.$14k - 28 = 0 \implies k = 2$.

Let $P$ be a plane $lx + my + nz = 0$ containing the line $\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$. If plane $P$ divides the line segment $AB$ joining points $A(-3, -6, 1)$ and $B(2, 4, -3)$ in the ratio $k : 1$,then the value of $k$ is equal to

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