The total number of $3\times 3$ matrices A having entries from the set $\left\{0,1,2,3\right\}$ such that the sum of all the diagonal entries of $A{A}^T$ is 9, is equal to _________

$A=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\Rightarrow A^T=\left[\begin{array}{ccc}a& d& g\\ b& e& h\\ c& f& i\end{array}\right]$

$A{A}^T=\left[\begin{array}{ccc}{a}^2+b^2+c^2& ad+be+cf& ag+bh+ci\\ da+eb+fc& d^2+e^2+f^2& dg+eh+fi\\ ga+hb+ic& gd+he+if& g^2+h^2+i^2\end{array}\right]$

Given, $a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2=9$

$a,b,c,d,e,f,g,h,i\in \left\{0,1,2,3\right\}$

(i) One of them is 3, remaining are 0s 9 ways 
(ii) 2 of them are 2s, one of them is 1 and remaining 0s $9C_2.7C_1=252$ ways
iii) One of them is 2’s five of them are 1’s and remaining 0’s $9_{C_1}{.8}_{C_5}=504$ ways 
(iv) All are 1s in 1 way
Total $=9+252+504+1=766$  ways