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 

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.

$\therefore D=0=D_1=D_2=D_3$

Now, $D=0\Rightarrow \left|\begin{array}{ccc}1& 4& -2\\ 1& 7& -5\\ 1& 5& \alpha \end{array}\right|=0$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(7\alpha +25)-4(\alpha +5)-2(5-7)=0\Rightarrow 3\alpha +9=0\Rightarrow \alpha =-3\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{D}_3=0\Rightarrow \left|\begin{array}{lll}1& 4& 1\\ 1& 7& \beta \\ 1& 5& 5\end{array}\right|=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(35-5\beta )-4(5-\beta )+2(5-7)=0\Rightarrow \beta =13\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\alpha +\beta =-3+13=10\end{array}$