Question

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

• A. $0$
• B. $10$
• C. $-10$
• D. $2$

Answer 1

The correct option is B $10$

The given planes intersect in a line
$\therefore D=D_x=D_y=D_z=0$
$D=0$
$\Rightarrow \left|\begin{array}{ccc}1& 4& -2\\ 1& 7& -5\\ 1& 5& \alpha \end{array}\right|=0$
$\Rightarrow 7\alpha +25-4\alpha -20+4=0\Rightarrow \alpha =-3$

$D_z=0$
$\left|\begin{array}{ccc}1& 4& 1\\ 1& 7& \beta \\ 1& 5& 5\end{array}\right|=0$
$\Rightarrow 35-5\beta -20+4\beta -2=0\Rightarrow \beta =13\therefore \alpha +\beta =10$