The sum of all the integral values of $p$ such that the equation $3\sin^2 x + 12\cos x - 3 = p$, $x \in \mathbb{R}$, has at least one solution, is:

Given equation: $3\sin^2 x + 12\cos x - 3 = p$

Substituting $\sin^2 x = 1 - \cos^2 x$:

$3(1 - \cos^2 x) + 12\cos x - 3 = p$

$-3\cos^2 x + 12\cos x = p$

Let $\cos x = t$. Since $x \in \mathbb{R}$, $t \in [-1, 1]$.

The equation becomes $p = -3t^2 + 12t$.

Let $f(t) = -3t^2 + 12t$.

Differentiating with respect to $t$:

$f'(t) = -6t + 12 = 6(2 - t)$

For $t \in [-1, 1]$, $f'(t) > 0$, which means $f(t)$ is strictly increasing in the interval $[-1, 1]$.

The minimum value of $f(t)$ occurs at $t = -1$:

$f(-1) = -3(-1)^2 + 12(-1) = -15$

The maximum value of $f(t)$ occurs at $t = 1$:

$f(1) = -3(1)^2 + 12(1) = 9$

Thus, the range of $p$ for which the equation has at least one solution is $[-15, 9]$.

The integral values of $p$ are $-15, -14, -13, \dots, 9$.

The sum of these integral values is:

$\sum_{k=-15}^{9} k = (-15) + (-14) + (-13) + (-12) + (-11) + (-10) + \sum_{k=-9}^{9} k$

Since $\sum_{k=-9}^{9} k = 0$, the sum simplifies to:

$-15 - 14 - 13 - 12 - 11 - 10 = -75$

Answer: $-75$