(A) Given the determinant $\Delta = \left|\begin{array}{ccc}a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z\end{array}\right|$.
Using the property of determinants,we can split the second row into two parts:
$\Delta = \left|\begin{array}{ccc}a & b & c \\ a & b & c \\ x & y & z\end{array}\right| + \left|\begin{array}{ccc}a & b & c \\ 2x & 2y & 2z \\ x & y & z\end{array}\right|$.
In the first determinant,the first row and second row are identical $(R_1 = R_2)$,so its value is $0$.
In the second determinant,we can factor out $2$ from the second row:
$\Delta = 0 + 2 \left|\begin{array}{ccc}a & b & c \\ x & y & z \\ x & y & z\end{array}\right|$.
Here,the second row and third row are identical $(R_2 = R_3)$,so its value is $0$.
Thus,$\Delta = 0 + 2(0) = 0$.