Explanation

(d)
General term in the expansion of
$(1 + 3x + {2x}^{2})^{6} = \sum_{}^{}{\frac{6!}{r_{1}!r_{2}!r_{3}!}(1)^{r_{1}}(3x)^{r_{2}}\left( {2x}^{2} \right)^{r_{3}}}$
Where $r_{1} + r_{2} + r_{3} = 6$ ….(i)
For coefficient of $x^{11}$, we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ r_{2} + 2\ r_{3} = 11$ …(ii)
Now, from Eqs, (i) and (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r_{1} = r_{3} - 5$
For $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r_{3} = 5,\ \ r_{1} = 0$
And $r_{2} = 1$
$\therefore Coefficient\ of\ x^{11} = \frac{6!}{0!1!5!}(1)^{0}(3)^{1}(2)^{5}$
$= 6 \times 3 \times 2^{5} = 18 \times 32 = 576$