If the lines frac{x-1}{-3}=frac{y-2}{2k}=frac | vedclass

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(A) The direction ratios of the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are $(-3, 2k, 2)

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$-\frac{10}{7}$

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(A) The direction ratios of the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are $(-3, 2k, 2)$ and $(3k, 1, -5)$ respectively.Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.Substituting the given values,we get:$-3(3k) + (2k)(1) + (2)(-5) = 0$$-9k + 2k - 10 = 0$$-7k - 10 = 0$$-7k = 10$$k = -\frac{10}{7}$Thus,the value of $k$ is $-\frac{10}{7}$.

If the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are perpendicular,find the value of $k$.

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