Magic square - 3.0

If the sum of three numbers in the same row/column/diagonal is $S$ , then

$(a+b+c)+(c+f+i)+(i+g+h)+(a+d+g)=4S=2a+2c+2g+2i+b+d+f+h=2(a+i)+2(c+g)+(b+h)+(d+f)=2(S-e)+2(S-e)+(S-e)+(S-e)=6(S-e)=6S-6e$

$\begin{aligned} 4S & = 6S-6e \\ 4S+6e & = 6S \\ 6e & = 2S \\ e & = \dfrac{2S}{6} \\ & =\dfrac{S}{3} \end{aligned}$

So $S=3*\dfrac{S}{3}=3*e=3*7=21=g+h+i$ .

Now we will prove that $a^2+b^2+c^2=g^2+h^2+i^2$ . We can use the figure below.

Now $a^2+b^2+c^2=a^2+b^2+(S-a-b)^2=S^2+2a^2+2b^2-2aS-2bS+2ab$ and $g^2+h^2+i^2=\left(a+b-\dfrac{S}{3}\right)^2+\left(\dfrac{2S}{3}-a\right)^2+\left(\dfrac{2S}{3}-b\right)^2=S^2+2a^2+2b^2-2aS-2bS+2ab$

So we proved that $a^2+b^2+c^2=g^2+h^2+i^2$ .

Now $\begin{aligned} (g+h+i)^2 & = 21^2 = 441 \\ & = g^2+h^2+i^2+2gh+2gi+2hi \\ & = a^2+b^2+c^2+2(gh+gi+hi) \\ & = 161+2(gh+gi+hi) \\ \ \\ \ \\ \Longrightarrow 2(gh+gi+hi) & = 441-161 \\ & = 280 \\ gh+gi+hi & =\dfrac{280}{2}=\boxed{140} \end{aligned}$

Note: Therefore the answer is $\boxed{140}$ , which is possible: