If for some alpha and beta in mathbb{R}, the | vedclass

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(A) For the three planes to intersect in a line,the system of linear equations must have infinitely many solutions. This occurs when the determinant o

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$10$

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(A) For the three planes to intersect in a line,the system of linear equations must have infinitely many solutions. This occurs when the determinant of the coefficient matrix $\Delta = 0$ and the augmented determinants $\Delta_x, \Delta_y, \Delta_z$ are also $0$.First,we calculate the determinant of the coefficient matrix:$\Delta = \begin{vmatrix} 1 & 4 & -2 \ 1 & 7 & -5 \ 1 & 5 & \alpha \end{vmatrix} = 1(7\alpha + 25) - 4(\alpha + 5) - 2(5 - 7) = 7\alpha + 25 - 4\alpha - 20 + 4 = 3\alpha + 9$.Setting $\Delta = 0$,we get $3\alpha + 9 = 0$,which implies $\alpha = -3$.Next,for the system to have infinitely many solutions,we must have $\Delta_z = 0$:$\Delta_z = \begin{vmatrix} 1 & 4 & 1 \ 1 & 7 & \beta \ 1 & 5 & 5 \end{vmatrix} = 1(35 - 5\beta) - 4(5 - \beta) + 1(5 - 7) = 35 - 5\beta - 20 + 4\beta - 2 = 13 - \beta$.Setting $\Delta_z = 0$,we get $13 - \beta = 0$,which implies $\beta = 13$.With $\alpha = -3$ and $\beta = 13$,the system is consistent and represents a line. Thus,$\alpha + \beta = -3 + 13 = 10$.

If for some $\alpha$ and $\beta$ in $\mathbb{R},$ the intersection of the following three planes $x+4y-2z=1$,$x+7y-5z=\beta$,and $x+5y+\alpha z=5$ is a line in $\mathbb{R}^{3},$ then $\alpha+\beta$ is equal to

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