Let a line having direction ratios 1, -4, 2 i | vedclass

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(D) Let point $A$ on the first line be $(3\lambda+7, -\lambda+1, \lambda-2)$ and point $B$ on the second line be $(2\mu, 3\mu+7, \mu)$.The direction r

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$84$

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(D) Let point $A$ on the first line be $(3\lambda+7, -\lambda+1, \lambda-2)$ and point $B$ on the second line be $(2\mu, 3\mu+7, \mu)$.The direction ratios of line $AB$ are $(2\mu - (3\lambda+7), 3\mu+7 - (-\lambda+1), \mu - (\lambda-2))$,which simplifies to $(2\mu - 3\lambda - 7, 3\mu + \lambda + 6, \mu - \lambda + 2)$.Since the direction ratios of the line $AB$ are given as $1, -4, 2$,we have:$\frac{2\mu - 3\lambda - 7}{1} = \frac{3\mu + \lambda + 6}{-4} = \frac{\mu - \lambda + 2}{2} = k$From the first and second ratios:$-8\mu + 12\lambda + 28 = 3\mu + \lambda + 6 \implies 11\lambda - 11\mu = -22 \implies \lambda - \mu = -2 \implies \mu = \lambda + 2$.From the second and third ratios:$6\mu + 2\lambda + 12 = -4\mu + 4\lambda - 8 \implies 10\mu - 2\lambda = -20 \implies 5\mu - \lambda = -10$.Substituting $\mu = \lambda + 2$ into $5\mu - \lambda = -10$:$5(\lambda + 2) - \lambda = -10 \implies 4\lambda + 10 = -10 \implies 4\lambda = -20 \implies \lambda = -5$.Then $\mu = -5 + 2 = -3$.Coordinates of $A$: $(3(-5)+7, -(-5)+1, -5-2) = (-8, 6, -7)$.Coordinates of $B$: $(2(-3), 3(-3)+7, -3) = (-6, -2, -3)$.$(AB)^2 = (-6 - (-8))^2 + (-2 - 6)^2 + (-3 - (-7))^2 = (2)^2 + (-8)^2 + (4)^2 = 4 + 64 + 16 = 84$.

Let a line having direction ratios $1, -4, 2$ intersect the lines $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}$ and $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}$ at the points $A$ and $B$ respectively. Then $( AB )^{2}$ is equal to

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