Let $$p, q \in Q$$. If $$2 - \sqrt{3}$$ is a root of the quadratic equation $$x^2 + px + q = 0$$, then:

We are given a quadratic equation with rational coefficients

$$x^2 + px + q \;=\; 0,$$

and we are told that $$2 - \sqrt{3}$$ is one of its roots. Because both $$p$$ and $$q$$ are rational numbers, any irrational part of a root must appear in conjugate form. Therefore the conjugate root must also be present:

$$2 + \sqrt{3}.$$

Now we invoke Vieta’s formulas, which state:

For a quadratic equation $$x^2 + px + q = 0,$$ if the roots are $$\alpha$$ and $$\beta,$$ then

$$\alpha + \beta \;=\; -p \quad\text{and}\quad \alpha \beta \;=\; q.$$

We substitute our two roots $$\alpha = 2 - \sqrt{3}$$ and $$\beta = 2 + \sqrt{3}.$$

First, the sum of the roots:

$$\alpha + \beta \;=\; (2 - \sqrt{3}) + (2 + \sqrt{3})$$

$$=\; 2 + 2 + (-\sqrt{3} + \sqrt{3})$$

$$=\; 4.$$

Using Vieta’s relation $$\alpha + \beta = -p,$$ we get

$$-p = 4 \quad\Longrightarrow\quad p = -4.$$

Next, the product of the roots:

$$\alpha \beta \;=\; (2 - \sqrt{3})(2 + \sqrt{3}).$$

We expand the product using the identity $$(a - b)(a + b) = a^2 - b^2$$ with $$a = 2$$ and $$b = \sqrt{3}:$$

$$\alpha \beta \;=\; 2^2 - (\sqrt{3})^2$$

$$=\; 4 - 3$$

$$=\; 1.$$

Using Vieta’s relation $$\alpha \beta = q,$$ we obtain

$$q = 1.$$

So we have found

$$p = -4 \quad\text{and}\quad q = 1.$$

Now we examine each given option by substituting these values.

Option A: $$p^2 - 4q + 12$$

$$=(-4)^2 - 4(1) + 12 = 16 - 4 + 12 = 24 \neq 0.$$

Option B: $$q^2 + 4p + 14$$

$$=(1)^2 + 4(-4) + 14 = 1 - 16 + 14 = -1 \neq 0.$$

Option C: $$p^2 - 4q - 12$$

$$=(-4)^2 - 4(1) - 12 = 16 - 4 - 12 = 0.$$

Option D: $$q^2 - 4p - 16$$

$$=(1)^2 - 4(-4) - 16 = 1 + 16 - 16 = 1 \neq 0.$$

Only Option C yields zero, satisfying the required condition.

Hence, the correct answer is Option C.