How many $3 \times 3$ matrices M with entries from $\{0, 1, 2\}$ are there, for which the sum of the diagonal entries of $M^{T} M$ is 5?

How many 3 × 3 matrices M with entries from {0, 1, 2} are there, for which the sum of the diagonal entries of M^{T} M is 5? $M \text{ is } 3\times3,\qquad m_{ij}\in\{0,1,2\}$ $\text{trace}\left(M^{T}M\right)=5$ $\text{To find: the number of such matrices } M$ $(1)\ 126 \quad (2)\ 198 \quad (3)\ 162 \quad (4)\ 135$ $(M^{T}M)_{ij}=\vec{c_i}\cdot\vec{c_j}$ $\text{trace}(M^{T}M)=|\vec{c_1}|^{2}+|\vec{c_2}|^{2}+|\vec{c_3}|^{2}$ $\vec{c_j}=\text{column } j \text{ of } M$ $\text{count ordered } (\vec{c_1},\vec{c_2},\vec{c_3}) \text{ with } \sum|\vec{c_j}|^{2}=5$ $\vec{c}\in\{0,1,2\}^{3}:\quad |\vec{c}|^{2}=a+4b\ \ (a=\#1,\ b=\#2)$ $|\vec{c}|^{2}\in\{0,1,2,3,4,5\}$ $N(0)=1,\ N(1)=3,\ N(2)=3,\ N(3)=1,\ N(4)=3,\ N(5)=6$ $(0,0,5),\quad (0,1,4),\quad (0,2,3),\quad (1,1,3),\quad (1,2,2)$ $(0,0,5):\ 3\cdot N(0)^2N(5)=3\cdot6=18$ $(0,1,4):\ 6\cdot N(0)N(1)N(4)=6\cdot9=54$ $(0,2,3):\ 6\cdot N(0)N(2)N(3)=6\cdot3=18$ $(1,1,3):\ 3\cdot N(1)^2N(3)=3\cdot9=27$ $(1,2,2):\ 3\cdot N(1)N(2)^2=3\cdot27=81$ $\text{Total}=18+54+18+27+81=198$ $198\ \Rightarrow\ \text{option }(2)$ $\boxed{198}\quad\Rightarrow\ (2)$