Explanation

(c)
We have, $x^{2} + 6x - 27 > 0$
$\Rightarrow (x + 9)(x - 3) > 0 \Rightarrow x < - 9\$or$\ x > 3$
$\Rightarrow x \in ( - \infty, - 9) \cup (3,\infty)\ \$….(i)
And $x^{2} - 3x - 4 < 0$
$\Rightarrow (x - 4)(x + 1) < 0$
$\Rightarrow - 1 < x < 4$ ….(i)
From relations (i) and (ii), we get
$3 < x < 4$