The set of all real values of p for which the equation $$3 \sin^{2}x + 12 \cos x - 3 = p$$ has at least one solution is

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

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

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

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

Adding and subtracting 12 in LHS

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

$$-3\left(\cos^2x-4\cos x+4\right)+12=p$$

$$-3\left(\cos x-2\right)^2+12=p\rightarrow1$$

Now, we know that -

$$-1\le\cos x\le1$$

$$-3\le\cos x-2\le-1$$

$$1\le\left(\cos x-2\right)^2\le9$$

$$-27\le-3\left(\cos x-2\right)^2\le-3$$

$$-15\le-3\left(\cos x-2\right)^2+12\le9$$

Substituting it as p, from eq. 1-

$$-15\le p\le9$$