The number of 3 × 2 matrices A, which can be formed using the...

For a 3×2 matrix A, the diagonal sum of $A^T A$ equals the sum of squares of all 6 elements of A. We need:
$a_{11}^2 + a_{21}^2 + a_{31}^2 + a_{12}^2 + a_{22}^2 + a_{32}^2 = 5$ where each element is from $\{-2, -1, 0, 1, 2\}$.
Possible squares: $0^2=0$, $(\pm 1)^2=1$, $(\pm 2)^2=4$.
Case 1:
Distribution $4+1+0+0+0+0$ (one ±2, one ±1, four 0's).
Arrangements: $\binom{6}{1} \times \binom{5}{1} = 30$.
Sign choices: $2 \times 2 = 4$. Total: $30 \times 4 = 120$.
Case 2:
Distribution $1+1+1+1+1+0$ (five ±1, one 0).
Arrangements: $\binom{6}{1} = 6$ ways to place the 0.
Sign choices for five ±1: $2^5 = 32$. Total: $6 \times 32 = 192$.
Total matrices: $120 + 192 = 312$