For (i) $16y^2 - 24y + 9$, we use the identity $(a + b)^2 = a^2 + 2ab + b^2$. By comparing, we find $a = 4y$ and $b = -3$. Substituting these values, we get $(4y)^2 + 2(4y)(-3) + (-3)^2 = \mathbf{(4y - 3)^2}$.

For (ii) $\frac{9}{4}s^2 + 6st + 4t^2$, we apply the identity $(a + b)^2 = a^2 + 2ab + b^2$. Here, $a = \frac{3}{2}s$ and $b = 2t$. This gives us $(\frac{3}{2}s)^2 + 2(\frac{3}{2}s)(2t) + (2t)^2 = \mathbf{(\frac{3}{2}s + 2t)^2}$.

For (iii) $\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2$, we use the identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$. Let $a = \frac{m}{3}$, $b = 3n$, and $c = \frac{k}{2}$. Expanding gives the exact expression, so it factors to $\mathbf{(\frac{m}{3} + 3n + \frac{k}{2})^2}$.

For (iv) $\frac{p^2}{16} - 2 + \frac{16}{p^2}$, we use the identity $(a + b)^2 = a^2 + 2ab + b^2$. Here, $a = \frac{p}{4}$ and $b = -\frac{4}{p}$. Substituting these, we get $(\frac{p}{4})^2 + 2(\frac{p}{4})(-\frac{4}{p}) + (-\frac{4}{p})^2 = \mathbf{(\frac{p}{4} - \frac{4}{p})^2}$.

For (v) $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc$, we use the identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$. Let $a = 3a$, $b = -2b$, and $c = c$. Expanding these terms matches the given expression, yielding the factors $\mathbf{(3a - 2b + c)^2}$.