Given ,
$A = \begin{bmatrix}x_1ㅤx_2ㅤ7 \\y_1ㅤy_2ㅤy_3 \\z_1ㅤ8ㅤ3 \end{bmatrix}$
Now lets assume the sum of any row or column or diagonals to be "K".
We can use the property of determinants C1 to C1+C2+C3 :
$A = \begin{bmatrix}Kㅤx_2ㅤ7 \\Kㅤy_2ㅤy_3 \\Kㅤ8ㅤ3 \end{bmatrix}$
Now apply R1 to R1+R2+R3 :
$A = \begin{bmatrix}3KㅤKㅤK \\Kㅤy_2ㅤy_3 \\Kㅤ8ㅤ3 \end{bmatrix}$
Now as we assumed : $K=7+3+Y_3$ , this implies $Y_3=K-10$ .... Equation 1
Now by equating Diagonal sum to C2 sum :$X_2+Y_2+8=X_1+Y_2+3$
This implies $X_1=X_2+5$
And we also know that sum of R1 to be K : $X_1+X_2+7=K$
Now by substituting $X_1$ in terms of $X_2$ : we get $X_2$= $\dfrac{\ K-12}{2}$
And we also know that Sum of C2 to be K : $X_2+Y_2+8=K$
By substituting $X_2$we get $Y_2=\dfrac{\ K-4}{2}$... Equation 2
Now we can substitute Equation 1 Equation 2 in the matrix :
$A = \begin{bmatrix}3KㅤKㅤK \\KㅤK/2-2ㅤK-10 \\Kㅤ8ㅤ3 \end{bmatrix}$
Now to find K:
We already know that Sum of R2 = Sum of C3 = K
$Y_1+Y_2+Y_3=10+Y_3$
Hence,$Y_1+Y_2=10$
We already deduced $Y_2$ = $\dfrac{\ K-4}{2}$
Hence, $Y_1=12-\dfrac{\ K}{2}$
Now , we also know that Sum of all elements should be 3(K) , and
This 3K should be equal to {Sum of two diagonals -$Y_2$ + 8 +$Y_1+X_2+Y_3$}
3K = 2K -$\dfrac{\ K-4}{2}$ + 8 + ($12-\dfrac{\ K}{2}$) + ($\dfrac{\ K-12}{2}$) + (K-10)
This implies K = 12.
Therefore the determinant of :
$A = \begin{bmatrix}3KㅤKㅤK \\KㅤK/2-2ㅤK-10 \\Kㅤ8ㅤ3 \end{bmatrix}$
$A = \begin{bmatrix}36ㅤ12ㅤ12 \\12ㅤ4ㅤ2 \\12ㅤ8ㅤ3 \end{bmatrix}$
= $36(-4)-12(12)+12(12(4))$
= $12\times12\times2$
= 288.
