Explanation

(b)
Any plane through (2, 2, 1) is
$a(x - 2) + b(y - 2) + c\left( \mathcal{z} - 1 \right) = 0$ (i)
It passes through (9, 3, 6) if $7a + b + 5c = 0$ (ii)
Also (i) is perpendicular to $2x + 6y + 6\mathcal{z} - 1 = 0$, we have
$2a + 6b + 6c = 0$
$\therefore\ a + 3b + 3c = 0$ (iii)
$\therefore\ \frac{a}{- 12} = \frac{b}{- 16} = \frac{c}{20}$ or $\frac{a}{3} = \frac{b}{4} = \frac{c}{- 5}$ (from (ii) and (iii))
Therefore, the required plane is $3(x - 2) + 4(y - 2) - 5\left( \mathcal{z}–1 \right) = 0\$or$\ 3x + 4y - 5\mathcal{z} - 9 = 0$