Given:-
- Height of cone = 7 units
- Radius of cone = 3 units
To find:
We know:-
[tex] \bigstar \boxed{ \rm Volume \: of \: Cone = \frac{1}{3} \pi {r}^{2} h}[/tex]
[tex] \\ \\ [/tex]
So:-
[tex] \leadsto\sf Volume \: of \: Cone = \dfrac{1}{3} \pi {r}^{2} h[/tex]
[tex] \\ \\ [/tex]
[tex] \leadsto\sf Volume \: of \: Cone = \dfrac{1}{3} \times \pi \times {3}^{2} \times 7 \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \leadsto\sf Volume \: of \: Cone = \dfrac{1}{3} \times \pi \times 3 \times 3 \times 7 \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \leadsto\sf Volume \: of \: Cone = \dfrac{1}{\cancel3} \times \pi \times \cancel3 \times 3 \times 7 \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \leadsto\sf Volume \: of \: Cone =\pi \times 3 \times 7 \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \leadsto\sf Volume \: of \: Cone =\pi \times 21\\ [/tex]
[tex] \\ \\ [/tex]
[tex] \leadsto\bf Volume \: of \: Cone = 21\pi[/tex]
[tex] \\ \\ [/tex]
Therefore option C is correct.