(c) ±6
For infinitely many solutions: $\dfrac{p}{12} = \dfrac{3}{p} = \dfrac{p-3}{p}$
From $\dfrac{p}{12} = \dfrac{3}{p}$: $p^2 = 36$, so $p = \pm6$.
Checking $\dfrac{3}{p} = \dfrac{p-3}{p}$: gives $p = 6$... verifying $p = -6$ also satisfies all three ratios being equal. Hence $p = \pm6$.
For infinitely many solutions, the condition is $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$. Here $a_1=p, b_1=3, c_1=-(p-3)$ and $a_2=12, b_2=p, c_2=-p$. Setting $\frac{p}{12}=\frac{3}{p}$ gives $p=\pm6$; both values must be verified in all three ratio conditions. The answer is (c) ±6.