7. Let $a, b, c, d$ all be prime numbers, satisfying: $a>3 b>6 c>12 d$, and
$$a^{2}-b^{2}+c^{2}-d^{2}=1749$$

Find the value of $a^{2}+b^{2}+c^{2}+d^{2}$.
7. From the conditions, we know that $a, b, c$ are all 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, and thus $d=2$. Consequently, $a^{2}-b^{2}+c^{2}=1753$. From the conditions, we also know that $a \geqslant 3 b+2, b \geqslant 2 c+1, c \geqslant 5$, hence
$$\begin{aligned}
1753 & \geqslant(3 b+2)^{2}-b^{2}+c^{2}=8 b^{2}+12 b+4-c^{2} \\
& \geqslant 8(2 c+1)^{2}+12 b+4=33 c^{2}+32 c+12 b+12 \\
& \geqslant 33 c^{2}+160+132+12
\end{aligned}$$

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