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

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(D) The given lines are $L_1: \frac{x-3}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}$ and $L_2: \frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5}$.The directio

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$-\frac{10}{7}$

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(D) The given lines are $L_1: \frac{x-3}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}$ and $L_2: \frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5}$.The direction ratios of $L_1$ are $\vec{a_1} = (-3, 2k, 2)$.The direction ratios of $L_2$ are $\vec{a_2} = (3k, 1, -5)$.Since the lines are perpendicular,the dot product of their direction ratios must be zero:$\vec{a_1} \cdot \vec{a_2} = 0$$(-3)(3k) + (2k)(1) + (2)(-5) = 0$$-9k + 2k - 10 = 0$$-7k = 10$$k = -\frac{10}{7}$.

If lines $\frac{x-3}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{6-z}{5}$ are perpendicular to each other,then $k=$ $\qquad$ .

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