Minimum

The area of the triangle is given by,

$S = \dfrac{1}{2} a b \sin \theta$

where $\theta$ is the angle between sides $a$ and $b$ . It follows that

$b = \dfrac{ 2 S }{ a \sin \theta}$

Also, we have from the Law of Cosines,

$c^2 = a^2 + b^2 - 2 a b \cos \theta = a^2 + \dfrac{ 4 S^2 }{a^2 \sin^2 \theta} - 4 S \cot \theta$

Therefore,

$f = a^2 + 4 b^2 + 12 c^2 = a^2 + \dfrac{16 S^2}{a^2 \sin^2 \theta} + 12 (a^2 + \dfrac{ 4 S^2 }{a^2 \sin^2 \theta} - 4 S \cot \theta)$

Simplifying,

$f = 13 a^2 + \dfrac{64 S^2 }{ a^2 \sin^2 \theta} - 48 S \cot \theta$

Differentiating $f$ with respect to $a^2$ and with respect to $\theta$ yields:

$\dfrac{\partial f}{\partial a^2} = 13 - \dfrac{64 S^2}{ \sin^2 \theta} (\dfrac{1}{a^4}) = 0$

From which, $a^2 = \dfrac{ 8 S }{ \sqrt{13} \sin \theta }$

And,

$\dfrac{\partial f}{\partial \theta} = \dfrac{ 64 S^2 }{ a^2 } (2 \cot \theta) (- \csc^2 \theta) - 48 S ( - csc^2 \theta ) = 0$

From which, $\dfrac{ 8 S }{a^2} \cot \theta = 3$ . Substituting this in the previous equation, we get

$\dfrac{8 S}{ a^2 } = \sqrt{13} \sin \theta = 3 \tan \theta$

Therefore, $\cos \theta = \dfrac{3}{\sqrt{13} }$ , and it follows from this that $\sin \theta = \dfrac{2}{\sqrt{13}}$

Therefore, $a^2 = \dfrac{ 8 S }{ \sqrt{13} \sin \theta } = 4 S$

Substituting this in the expression for the function $f$ ,

$\begin{aligned} f &= 13 a^2 + \dfrac{64 S^2 }{ a^2 \sin^2 \theta} - 48 S \cot \theta \\ &= 13 (4 S) + 16 S \cdot (\dfrac{13}{4} ) - 48 S \cdot (\dfrac{3}{2}) \\ &= 52 S + 52 S - 72 S = 104 S - 72 S = 32 S \end{aligned}$

Therefore, the answer is $\boxed{32}$