Let the three numbers be $x,y$ and $z.$
Let, ashok took $z = 73 $ instead of $ z = 37 $
So, the final product will be
$ xy(73) – xy(37) = 720 $
$ \Rightarrow xy(73 -37) = 720 $
$ \Rightarrow xy(36) = 720 $
$ \Rightarrow \boxed {xy = 20} \quad \longrightarrow (1)$
We know that, $\text{AM}$ and $\text{GM}$ are arithmetic and geometric mean respectively between two numbers $x$ and $y$, then
$ \boxed {\text{AM} \geqslant \text{GM}}$
$ \Rightarrow \frac{(x+y)}{2} \geqslant \sqrt{xy}$
$ \Rightarrow x+y \geqslant 2 \sqrt{xy} $
$ \Rightarrow (x+y)^{2} \geqslant 4xy$
$ \Rightarrow x^{2} + y^{2} + 2xy \geqslant 4xy $
$ \Rightarrow x^{2} + y^{2} \geqslant 2xy $
$ \Rightarrow x^{2} + y^{2} \geqslant 2(20) $
$ \Rightarrow \boxed {x^{2} + y^{2} \geqslant 40} $
$\textbf{Short Method:}$
The minimum possible sum of the squares of the two numbers when $ x = y. $
From equation $ (1), x = y = \sqrt {20}$
so, $ x^{2} + y^{2} = (\sqrt {20})^{2} + ( \sqrt {20})^{2} $
$ \Rightarrow x^{2} + y^{2} = 20 + 20 = 40 $
$\therefore$ The minimum possible value of the sum of squares of the other two numbers is $40.$
Correct Answer $:40$