Matrices and Determinants Part 1

For the three planes to intersect in a line, the system must have infinitely many solutions with one free parameter, so the coefficient determinant must vanish.$\Delta=\begin{vmatrix}1&4&-2\\1&7&-5\\1&5&\alpha\end{vmatrix}=0$After subtracting row 1 from rows 2 and 3, this determinant becomes:$\Delta=\begin{vmatrix}1&4&-2\\0&3&-3\\0&1&\alpha+2\end{vmatrix}=3(\alpha+2)+3=3\alpha+9$Thus $3\alpha+9=0$, so $\alpha=-3$.For consistency, the Cramer determinant obtained by replacing the $z$-column with the constants must also vanish:$\Delta_z=\begin{vmatrix}1&4&1\\1&7&\beta\\1&5&5\end{vmatrix}=0$Using row operations, this becomes:$\Delta_z=\begin{vmatrix}1&4&1\\0&3&\beta-1\\0&1&4\end{vmatrix}=12-(\beta-1)=13-\beta$Hence $13-\beta=0$, so $\beta=13$.With $\alpha=-3$ and $\beta=13$, the three planes form a consistent rank $2$ system whose solution set is a line.Therefore, $\alpha+\beta=-3+13=10$.The correct option is $10$.