If $A=\left[\begin{array}{ccc}a& x& y\\ x& b& z\\ y& z& c\end{array}\right]$  where $a,b,c,x,y,z\in \{1,2,3,4,5,6\}$ and also $a,b,c,x,y,z$ are distinct then number of in A with trace equal to 10 are: 

$\because T_r\left(A\right)=a+b+c=10;(a\ne b\ne c)$

Therefore it may b  $(1,3,6),(1,4,5),(2,3,5)$
 $\therefore$ number of matrices = 3!$\times$ 3 $\times$  3!
Arranging diagonal  arranging non–diagonal elements
Elements
 $=3{(3!)}^2$