If  the  system  of  equations x + y + z = 5, x + 2y + 3z = 9, $x+3y+\alpha z=\beta$ has  infinitely  many solutions, then $\beta -\alpha$  equals

Since, the system of equations has infinitely many Solution, therefore $D=D_1=D_2=D_3=0.$ Here, 

 $D=\left|\begin{array}{l}111\\ 123\\ 13\alpha \end{array}\right|=1(2\alpha -9)-1(\alpha -3)+1(3-2)$ And $D_3=\left|\begin{array}{l}111\\ 123\\ 13\beta \end{array}\right|=1(2\beta -27)-1(\beta -9)+1(3-2)$

$=\beta -13$

Now ,        D = 0 $\Rightarrow \alpha -5=0\Rightarrow \alpha =5$ And $D_3=0\Rightarrow \beta -13=0$ $\Rightarrow \beta -13$

$\therefore \beta -\alpha =13-5=8$