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

We have two lines in symmetric form, and we need to find the value of $k$ that makes them perpendicular.

Two lines are perpendicular when their direction vectors are perpendicular, which means their dot product equals zero.

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Extract the direction ratios from the first line: $\frac{1-x}{3} = \frac{3y-6}{k} = \frac{3-z}{-2}$

Rewrite each term in the form $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$:

$\frac{1-x}{3} = \frac{-(x-1)}{3} = \frac{x-1}{-3}$

$\frac{3y-6}{k} = \frac{3(y-2)}{k} = \frac{y-2}{k/3}$

$\frac{3-z}{-2} = \frac{-(z-3)}{-2} = \frac{z-3}{2}$

Direction ratios of Line 1: $\left(-3, \frac{k}{3}, 2\right)$

Multiply by 3 to avoid fractions: $(-9, k, 6)$ or equivalently $(9, -k, -6)$

So $\vec{d_1} = (9, -k, -6)$

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Extract the direction ratios from the second line: $\frac{1-x}{2k} = \frac{y-5}{3} = \frac{6-z}{5}$

Rewrite:

$\frac{1-x}{2k} = \frac{x-1}{-2k}$

$\frac{y-5}{3}$

$\frac{6-z}{5} = \frac{z-6}{-5}$

Direction ratios of Line 2: $\vec{d_2} = (-2k, 3, -5)$

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Apply the perpendicularity condition: $\vec{d_1} \cdot \vec{d_2} = 0$

$(9)(-2k) + (-k)(3) + (-6)(-5) = 0$

$-18k - 3k + 30 = 0$

$-21k + 30 = 0$

$-21k = -30$

$k = \frac{30}{21} = \frac{10}{7}$

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$k = \frac{10}{7}$