Explanation

(c)
Given one vertex $A(7,\ 2,\ 4)$ and line $\frac{x + 6}{5} = \frac{y + 10}{3} = \frac{\mathcal{z} + 14}{8}$
General point on above line $B \equiv (5\lambda - 6,\ 3\lambda - 10,\ 8\lambda - 14)$
Direction ratios of line $AB$ are $< 5\lambda - 13,\ 3\lambda - 12,\ 8\lambda - 18 >$
Direction ratios of line $BC$ are <$5,\ 3,\ 8$>
Since angle between $AB$ nad $BC$ is $\pi/4$
$\cos\frac{\pi}{4} = \frac{(5\lambda - 3)5 + 3(3\lambda - 12) + 8(8\lambda - 18)}{\sqrt{5^{2} + 3^{2} + 8^{2}} \bullet \sqrt{(5\lambda - 13)^{2} + (3\lambda - 12)^{2} + (8\lambda - 18)^{2}}}$
Squaring and solving, we have $\lambda = 3,\ 2$
Hence equation of lines are $\frac{x - 7}{2} = \frac{y - 2}{- 3} = \frac{\mathcal{z} - 4}{6}$ and $\frac{x - 7}{3} = \frac{y - 2}{6} = \frac{\mathcal{z} - 4}{2}$