$\text{ The total number of }3\times 3\text{ matrices A having entries from the set }\{0,1,2,3\}\text{ such that the sum of }$

$\text{ all the diagonal entries of }A{A}^T\text{ is }9\text{ , is equal to }$

$\begin{array}{l}A=\left[\begin{array}{lll}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\Rightarrow A^T=\left[\begin{array}{lll}a& d& g\\ b& e& h\\ c& f& i\end{array}\right]\\ A{A}^T=\left[\begin{array}{lll}{a}^2+b^2+c^2& ad+be+cf ag+bh+ci& \\ da+eb+fc& d^2+e^2+f^2& dg+eh+fi\\ ga+hb+ic& gd+he+if& g^2+h^2+i^2\end{array}\right]\\ \text{ Given, }{a}^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2=9\text{ and }a,b,c,d,e,f,g,h,i\in \{0,1,2,3\}\end{array}$

Case(i)  One of them is 3, remaining are 0s is 9 ways

$\text{ Case (ii) }2\text{ of them are }2 \mathrm{s}\text{ , one of them is }1\text{ and remaining }0\text{ s is }{9}_{c_2}·7_{C_1}=252\text{ ways }$

$\text{ Case (iii) One of them is }2\text{ , five of them are }1\text{ s and remaining }0\text{ s is }{9}_{c_1}·8_{c_5}=504\text{ Ways }$

$\begin{array}{l}\text{Case }\left(iv\right)\text{\hspace{0.17em}All are 1s in one way}\\ \\ \text{Total: }9+252+504+1=767\text{\hspace{0.17em}ways}\end{array}$