The area of a triangle whose side lengths are $a, b,$ and $c$ is given by
$$A = \sqrt{s(s-a)(s-b)(s-c)} \quad \text{(Heron's formula)}$$
Where, $s = \dfrac{\text{(Perimeter of the triangle)}}{2} = \dfrac{a + b + c}{2} = \text{semi-perimeter of the triangle}$
Now, $a = 7, b = 8,$ and $c = 9$
$s = \dfrac{7 + 8 + 9}{2} = \dfrac{24}{2} = 12.$
Now, $A = \sqrt{12(12-7)(12-8)(12-9)} = \sqrt{12(5)(4)(3)} = \sqrt{4^{2} \times 3^{2} \times 5 } = 12\sqrt{5}$ sq. units.
So, the correct answer is $(B).$