If $$A = \begin{bmatrix}x_1 & x_2 & 7 \\y_1 & y_2 & y_3 \\z_1 & 8 & 3 \end{bmatrix}$$ is a matrix such that the sum of all three elements along any row, column or diagonal are equal to each other, then the value of determinant of A is:

Given ,
$$A = \begin{bmatrix}x_1 & x_2 & 7 \\y_1 & y_2 & y_3 \\z_1 & 8 & 3 \end{bmatrix}$$
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 :
$$A = \begin{bmatrix}K & x_2 & 7 \\K & y_2 & y_3 \\K & 8 & 3 \end{bmatrix}$$

Now apply R1 to R1+R2+R3 :
$$A = \begin{bmatrix}3K & K & K \\K & y_2 & y_3 \\K & 8 & 3 \end{bmatrix}$$

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 :
$$A = \begin{bmatrix}3K & K & K \\K & K/2-2 & K-10 \\K & 8 & 3 \end{bmatrix}$$

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 :
$$A = \begin{bmatrix}3K & K & K \\K & K/2-2 & K-10 \\K & 8 & 3 \end{bmatrix}$$

$$A = \begin{bmatrix}36 & 12 & 12 \\12 & 4 & 2 \\12 & 8 & 3 \end{bmatrix}$$

= $$36(-4)-12(12)+12(12(4))$$
= $$12\times12\times2$$
= 288.