The value of $p$ so that the lines $\frac{x-1}{-3}=\frac{2y-2}{2p}=\frac{z-3}{2}$ and $\frac{x-1}{-3p}=\frac{y-1}{4}=\frac{6-z}{5}$ are at right angles is

$\frac{10}{13}$

The correct answer is Option (3) → $\frac{10}{13}$

Direction ratios of line (1):

$\frac{x-1}{-3} = \frac{2y-2}{2p} = \frac{z-3}{2}$

⇒ direction ratios = $(-3,\,p,\,2)$

Direction ratios of line (2):

$\frac{x-1}{-3p} = \frac{y-1}{4} = \frac{6-z}{5}$

⇒ direction ratios = $(-3p,\,4,\,-5)$

If the lines are perpendicular, then their dot product = 0:

$(-3)(-3p) + (p)(4) + (2)(-5) = 0$

$9p + 4p - 10 = 0$

$13p - 10 = 0$

$p = \frac{10}{13}$

Final Answer: $p = \frac{10}{13}$