The radius of the base of a cone is increasing at the rate $3 \mathrm{~cm} /$ minute and the altitude is decreasing at the rate $4 \mathrm{~cm} /$ minute. The rate at which the lateral surface area is changing, when the radius is 7 cm and altitude is 24 cm . is

Given a right circular cone with base radius $r$, altitude $h$, and slant height $l = \sqrt{r^2 + h^2}$, the lateral surface area is $S = \pi r l$.

Differentiating with respect to time $t$:

$\frac{dS}{dt} = \pi \left[ \frac{dr}{dt} l + r \cdot \frac{1}{2\sqrt{r^2 + h^2}} \left( 2r \frac{dr}{dt} + 2h \frac{dh}{dt} \right) \right]$

At the instant $r = 7$ cm, $h = 24$ cm, the slant height is $l = \sqrt{7^2 + 24^2} = 25$ cm. With $\frac{dr}{dt} = 3$ cm/min and $\frac{dh}{dt} = -4$ cm/min:

$\frac{dS}{dt} = \pi \left[ 3 \cdot 25 + \frac{7}{25} \left( 7 \cdot 3 + 24 \cdot (-4) \right) \right] = \pi \left[ 75 + \frac{7}{25} (21 - 96) \right]$

$\frac{dS}{dt} = \pi \left[ 75 + \frac{7}{25} (-75) \right] = \pi (75 - 21) = 54\pi$ cm²/min

Final answer: $\boxed{\text{D}}$