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

Two lines in 3D space are perpendicular when their direction vectors are perpendicular to each other.

When direction vectors $\vec{d_1}$ and $\vec{d_2}$ are perpendicular:

$\vec{d_1} \cdot \vec{d_2} = 0$

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Given: $\frac{1-x}{3} = \frac{y-2}{2\lambda} = \frac{z-3}{2}$

Rewriting the first term:

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

Line 1 becomes: $\frac{x-1}{-3} = \frac{y-2}{2\lambda} = \frac{z-3}{2}$

Direction vector of Line 1: $\vec{d_1} = (-3, 2\lambda, 2)$

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Given: $\frac{x-1}{3\lambda} = \frac{y-1}{1} = \frac{6-z}{5}$

Rewriting the z-term:

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

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

Direction vector of Line 2: $\vec{d_2} = (3\lambda, 1, -5)$

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Since the lines are perpendicular:

$\vec{d_1} \cdot \vec{d_2} = 0$

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

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

$-7\lambda - 10 = 0$

$-7\lambda = 10$

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

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