A vertical pole of height h stands at the centroid of a triangle...

Let the centroid of the triangle $ABC$ be $G$. The vertical pole is positioned at $G$ and has height $h$.

The horizontal distances from the centroid to the vertices can be expressed using the angles of elevation:
$AG = h \cot \alpha$
$BG = h \cot \beta$
$CG = h \cot \gamma$

The distance from a vertex to the centroid is two-thirds the length of the corresponding median. For median $m_a$ to side $a$:
$AG = \dfrac{2}{3}m_a \Rightarrow AG^2 = \dfrac{4}{9}m_a^2$

Using Apollonius's theorem, $m_a^2 = \dfrac{1}{4}(2b^2 + 2c^2 - a^2)$. Thus,
$AG^2 = \dfrac{1}{9}(2b^2 + 2c^2 - a^2)$

Similarly, for the other vertices:
$BG^2 = \dfrac{1}{9}(2a^2 + 2c^2 - b^2)$
$CG^2 = \dfrac{1}{9}(2a^2 + 2b^2 - c^2)$

Adding these three squared distances gives:
$AG^2 + BG^2 + CG^2 = \dfrac{1}{9}(3a^2 + 3b^2 + 3c^2) = \dfrac{1}{3}(a^2+b^2+c^2)$

Rearranging for the sum of the squares of the sides:
$a^2+b^2+c^2 = 3(AG^2 + BG^2 + CG^2)$

Substituting the cotangent expressions:
$a^2+b^2+c^2 = 3(h^2 \cot^2 \alpha + h^2 \cot^2 \beta + h^2 \cot^2 \gamma)$

$\Rightarrow \dfrac{a^2+b^2+c^2}{h^2} = 3(\cot^2 \alpha + \cot^2 \beta + \cot^2 \gamma)$

Answer: $3(\cot^2 \alpha + \cot^2 \beta + \cot^2 \gamma)$