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

The lines need to be rewritten in the standard $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ form to extract direction ratios.

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

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

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

Direction ratios of Line 1: $(-3,\; p,\; 2)$

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Line 2: $\frac{x-1}{-3p} = \frac{y-1}{4} = \frac{6-z}{5}$

Since $6 - z = -(z - 6)$:

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

Direction ratios of Line 2: $(-3p,\; 4,\; -5)$

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Two lines are perpendicular when the dot product of their direction ratios equals zero:

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

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

$9p + 4p - 10 = 0$

$13p = 10$

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