If planes $x+4y -2z=1, x+7y -5z = \beta $ and $x+5y + \alpha z= 5 $ intersect in a line in $R^3$, then $\alpha + \beta $ is equal to

10

The correct answer is option (2) : 10

Given planes intersect in a line. This means that the system of equations $x+ 4y -2z=1, x+ 7y - 5z = \beta $ and $x+ 5y + \alpha z = 5 $ has infinitely many solutions.

$∴D= 0 = D_1= D_2= D_3 $

Now, $D=0 ⇒\begin{vmatrix}1 & 4 & -2\\1 & 7 & -5\\1 & 5 & \alpha \end{vmatrix}=0$

$⇒(7\alpha + 25) - 4(\alpha + 5) - 2( 5-7) = 0 $

$⇒3\alpha + 9 = 0 ⇒ \alpha = - 3$

$D_3= 0 ⇒\begin{vmatrix}1 & 4 & 1\\1 & 7 & \beta \\1 & 5 & 5\end{vmatrix}=0$

$⇒\alpha + \beta = -3+ 13= 10 $