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

$\frac{70}{11}$

The correct answer is Option (1) → $\frac{70}{11}$

Given lines:

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

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

Direction ratios of the first line:

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

So direction ratios are:

$\langle -3,\; \frac{2\lambda}{7},\; 2\rangle$

Direction ratios of the second line:

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

So direction ratios are:

$\langle -\frac{3\lambda}{7},\; 1,\; -5\rangle$

Lines are perpendicular ⇒ dot product = 0:

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

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

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

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

Final Answer: $\lambda = \frac{70}{11}$