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(A) The given system of equations is:$x+2y+z=1$$x+3y+4z=k$$x+5y+10z=k^2$For the system to be consistent,the determinant of the coefficient matrix $D$

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$3$

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(A) The given system of equations is:$x+2y+z=1$$x+3y+4z=k$$x+5y+10z=k^2$For the system to be consistent,the determinant of the coefficient matrix $D$ must be $0$ if the system has infinitely many solutions or a unique solution. Let us calculate $D$:$D = \begin{vmatrix} 1 & 2 & 1 \ 1 & 3 & 4 \ 1 & 5 & 10 \end{vmatrix} = 1(30-20) - 2(10-4) + 1(5-3) = 10 - 12 + 2 = 0$.Since $D=0$,for the system to be consistent,we must have $D_1 = D_2 = D_3 = 0$. Let us calculate $D_1$:$D_1 = \begin{vmatrix} 1 & 2 & 1 \ k & 3 & 4 \ k^2 & 5 & 10 \end{vmatrix} = 1(30-20) - 2(10k-4k^2) + 1(5k-3k^2) = 0$.$10 - 20k + 8k^2 + 5k - 3k^2 = 0$.$5k^2 - 15k + 10 = 0$.Dividing by $5$,we get $k^2 - 3k + 2 = 0$.$(k-1)(k-2) = 0$.Thus,the real values of $k$ are $k=1$ and $k=2$. Let $A=1$ and $B=2$.Therefore,$A+B = 1+2 = 3$.

If $A$ and $B$ are the two real values of $k$ for which the system of equations $x+2y+z=1$,$x+3y+4z=k$,and $x+5y+10z=k^2$ is consistent,then $A+B=$

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