Solve this following

In given question $\mathrm{p}, \mathrm{q} \in \mathrm{R}$. If we take other root as any real number $\alpha$, then quadratic equation will be

$x^{2}-(\alpha+2-\sqrt{3}) x+\alpha \cdot(2-\sqrt{3})=0$

Now, we can have none or any of the options can be correct depending upon ' $\alpha$ ' Instead of $\mathrm{p}, \mathrm{q} \in \mathrm{R}$ it should be $\mathrm{p}, \mathrm{q} \in \mathrm{Q}$ then other root will be $2+\sqrt{3}$

$\Rightarrow \mathrm{p}=-(2+\sqrt{3}-2-\sqrt{3})=-4$

and $\mathrm{q}=(2+\sqrt{3})(2-\sqrt{3})=1$

$\Rightarrow \mathrm{p}^{2}-4 \mathrm{q}-12=(-4)^{2}-4-12$

$=16-16=0$

Option (2) is correct