The value of $\lambda$ so that the lines $\frac{1-x}{3} = \frac{7y-14}{2\lambda} = \frac{z-3}{2}$ and $\frac{7-7x}{3\lambda} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angle, is:

The standard form of a line in 3D is:

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

where $(a, b, c)$ are the direction ratios of the line.

Line 1: $\frac{1-x}{3} = \frac{7y-14}{2\lambda} = \frac{z-3}{2}$

$\frac{-(x-1)}{3} = \frac{7(y-2)}{2\lambda} = \frac{z-3}{2}$

$\frac{x-1}{-3} = \frac{y-2}{\;\tfrac{2\lambda}{7}\;} = \frac{z-3}{2}$

Direction ratios of Line 1: $\;a_1 = -3,\quad b_1 = \dfrac{2\lambda}{7},\quad c_1 = 2$

Line 2: $\frac{7-7x}{3\lambda} = \frac{y-5}{1} = \frac{6-z}{5}$

$\frac{-7(x-1)}{3\lambda} = \frac{y-5}{1} = \frac{-(z-6)}{5}$

$\frac{x-1}{\;\tfrac{-3\lambda}{7}\;} = \frac{y-5}{1} = \frac{z-6}{-5}$

Direction ratios of Line 2: $\;a_2 = \dfrac{-3\lambda}{7},\quad b_2 = 1,\quad c_2 = -5$

Two lines are perpendicular when:

$a_1\,a_2 + b_1\,b_2 + c_1\,c_2 = 0$

$(-3)\!\left(\frac{-3\lambda}{7}\right) + \left(\frac{2\lambda}{7}\right)\!(1) + (2)(-5) = 0$

$\frac{9\lambda}{7} + \frac{2\lambda}{7} - 10 = 0$

$\frac{11\lambda}{7} = 10$

$\lambda = \dfrac{70}{11}$