Question:
If for some $\alpha$ and $\beta$ in $\mathbf{R}$, the intersection of the following three planes
$x+4 y-2 z=1$
$x+7 y-5 z=\beta$
$x+5 y+\alpha z=5$
is a line in $R^{3}$, then $\alpha+\beta$ is equal to:
Correct Option: , 2
Solution:
$\Delta=0 \Rightarrow\left|\begin{array}{ccc}1 & 4 & -2 \\ 1 & 7 & -5 \\ 1 & 5 & \alpha\end{array}\right|=0$
$\Rightarrow(7 \alpha+25)-(4 \alpha+10)+(-20+14)=0$
$\Rightarrow 3 \alpha+9=0 \Rightarrow \alpha=-3$
Also, $D_{z}=0 \Rightarrow\left|\begin{array}{lll}1 & 4 & 1 \\ 1 & 7 & \beta \\ 1 & 5 & 5\end{array}\right|=0$
$\Rightarrow 1(35-5 \beta)-(15)+1(4 \beta-7)=0 \Rightarrow \beta=13$
Hence, $\alpha+\beta=-3+13=10$