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

$\frac{10}{7}$

The correct answer is Option (1) → $\frac{10}{7}$

Given lines:

$\frac{1-x}{3}=\frac{3y-6}{k}=\frac{3-z}{-2}$

$\frac{1-x}{2k}=\frac{y-5}{3}=\frac{6-z}{5}$

Convert to symmetric form.

Line $L_{1}$ direction ratios:

$\frac{x-1}{-3}=\frac{y-2}{\frac{k}{3}}=\frac{z-3}{2}$

So DRs of $L_{1}$ are $(-3,\frac{k}{3},2)$.

Line $L_{2}$ direction ratios:

$\frac{x-1}{-2k}=\frac{y-5}{3}=\frac{z-6}{-5}$

So DRs of $L_{2}$ are $(-2k,3,-5)$.

Since lines are perpendicular, dot product of DRs is zero.

$(-3)(-2k)+\frac{k}{3}\cdot 3 + 2(-5)=0$

$6k + k - 10 = 0$

$7k = 10$

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

Final answer: $k=\frac{10}{7}$