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I think I get the logic of it but I don't have a proof.
A = G+H+IF = B+E+IJ = B+C+G
That means they are each at least 6 and none of B, C, E, G, H, or I can be larger than 6, which means they make up 1-6 in some order.
Then, it must be impossible to arrange them such that any of A, F, or G are equal to seven without one of them being greater than 10.
That leaves 7 for D.
Yes, 7 is correct: #731084
I think I get the logic of it but I don't have a proof.
A = G+H+I
F = B+E+I
J = B+C+G
That means they are each at least 6 and none of B, C, E, G, H, or I can be larger than 6, which means they make up 1-6 in some order.
Then, it must be impossible to arrange them such that any of A, F, or G are equal to seven without one of them being greater than 10.
That leaves 7 for D.