10. (10 points) Given 99 distinct prime numbers $p_{1}, p_{2}, \cdots p_{99}$, let $N=\mathrm{p}_{1}{ }^{2}+\mathrm{p}_{2}{ }^{2}+\cdots+\mathrm{p}_{99}{ }^{2}$, what is the remainder when $N$ is divided by 3?
【Analysis】Except for 3, the remainder of a prime number divided by 3 can only be 1 or 2, and the remainder of the square of a prime number divided by 3 can only be 1, 
(2 squared divided by 3 leaves a remainder of 1), then discuss whether it contains the prime number 3.
【Solution】Solution: Except for 3, the remainder of a prime number divided by 3 can only be 1 or 2, and the remainder of the square of a prime number divided by 3 can only be 
1, 
so 99 remainders of 1 add up to 99, and when divided by 3, the remainder is 0;
If these prime numbers include 3, since $3^{2} \div 3=3$, the remainder is 0, 
so 99 remainders add up to 98, and $98 \div 3=32 \cdots 2$,
Answer: The remainder of $N$ divided by 3 is 0 or 2.
Therefore, the answer is: 0 or 2.