Given: The mean of three numbers $2x$, $3x$, $x^2$ is $12$

So we can write: $\frac{2x + 3x + x^2}{3} = 12$

Simplify the numerator: $\frac{x^2 + 5x}{3} = 12$

Multiply both sides of the equation by $3$ to get: $x^2 + 5x = 36$

Subtract $36$ from both sides: $x^2 + 5x - 36 = 0$

Factor: $(x + 9)(x - 4) = 0$

So, either $x = -9$ or $x = 4$

Since we are told $x<0$, we can be certain that $x = -9$

If $x = -9$, then the values of the three numbers are as follows:

$2x = 2(-9) = -18$
$3x = 3(-9) = -27$
$x^2 = (-9)^2 = 81$

Range = (largest number) - (smallest number)
$= 81 - (-27)$
$= 108$

Answer: A