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

$\frac{-10}{7}$

The correct answer is Option (2) → $\frac{-10}{7}$

The given lines are:

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

For the first line, the direction ratios are obtained from the denominators:

$\Rightarrow$ direction ratios of line 1: $(-3,\;2\lambda,\;2)$

For the second line, rewriting $\frac{6-z}{5}$ as $\frac{z-6}{-5}$ gives direction ratios:

$\Rightarrow$ direction ratios of line 2: $(3\lambda,\;1,\;-5)$

Since the lines are perpendicular, their dot product is zero:

$(-3)(3\lambda)+(2\lambda)(1)+(2)(-5)=0$

$-9\lambda+2\lambda-10=0$

$-7\lambda-10=0$

Hence,

$\lambda=-\frac{10}{7}$