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

$(A)\; \lambda = -\frac{70}{11}$
$(B)\; \lambda = \frac{66}{11}$
$(C)\; \lambda = \frac{70}{11}$
$(D)\; \lambda = -\frac{66}{11}$

Let $L_1 = \frac{x-1}{-3} = \frac{y-2}{\frac{2 \lambda}{7}} = \frac{z-3}{2}$
and $L_2 = \frac{x-1}{-\frac{3 \lambda}{7}} = \frac{y-5}{1} = \frac{z-6}{-5}$
$-3 \times \frac{-3 \lambda}{7} + \frac{2\lambda}{7} \times 1 + 2 \times -5 = 0$
$\frac{9 \lambda}{7} + \frac{2 \lambda}{7} -10 = 0$
$\implies \frac{11 \lambda}{7} = 10$
$\implies \lambda = \frac{70}{11}$