7. Let $a, b, c, d$ all be prime numbers, and $a>3b>6c>12d, a^{2}-b^{2}+c^{2}-d^{2}=1749$. Find all possible values of $a^{2}+$ $b^{2}+c^{2}+d^{2}$.
7. From the conditions, $a, b, c$ are odd numbers. If $d$ is odd, then $a^{2}-b^{2}+c^{2}-d^{2}$ is even, which is a contradiction. Therefore, $d$ is even, so $d=2$, and thus $a^{2}-b^{2}+c^{2}=1753$. From the conditions, we also know that $a \geqslant 3 b+2, b \geqslant 2 c+1, c \geqslant 5$.
Hence, $1753 \geqslant(3 b+2)^{2}-b^{2}+c^{2}=8 b^{2}+12 b+4-c^{2} \geqslant 8(2 c+1)^{2}+12 b+4$
$$
=33 c^{2}+32 c+12 b+12 \geqslant 33 c^{2}+160+132+12 \text {, }
$$

Therefore, $c^{2}3 b$, we know $a-b>2 b \geqslant 22$, and both $a-b$ and $a+b$ are even, and $a, b$ are both odd, so $a-b=32, a+b=54$, hence $a=43, b=11$, and thus $a^{2}+b^{2}+c^{2}+d^{2}=1999$.