If $a,b,c\in N$ , the probability that   $a^2+b^2+c^2=13k,k\in N$ is $\frac{m}{n}$, where $m,n$  are relatively prime numbers, then  $(n-m)$ is

 $a,b,c\in N$
The remainder when  $a^2,b^2,c^2$  is divided by 13 is  $\{0,1,3,4,9,10,12\}$
Let E be the event that  $a^2+b^2+c^2=13k$
$\Rightarrow P\left(E\right)=\frac{1+3\times 1\times 2\times 2\times 3!+2\times 2\times 2\times 3!+2\times 2\times 2\times 3!}{{13}^3}=\frac{169}{{13}^3}=\frac{1}{13}$

$\Rightarrow 13-1=12$