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,

$\begin{array}{l}D=\left|\begin{array}{lll}1& 1& 1\\ 1& 2& 3\\ 1& 3& \alpha \end{array}\right|=1(2\alpha -9)-1(\alpha -3)+1(3-2)\\ =\alpha -5\end{array}$

$\text{ And }{D}_3=\left|\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}5\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}9\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\beta \end{array}\right|=1(2\beta -27)-1(\beta -9)+5(3-2)$

and $D_3=\left|\begin{array}{ccc}1& 1& 5\\ 1& 2& 9\\ 1& 3& \beta \end{array}\right|=1\left(2\beta -27\right)-1\left(\beta -9\right)+5\left(3-2\right)$

$=\beta -13$

$\text{ Now, }D=0$

$\Rightarrow \alpha -5=0\Rightarrow \alpha =5$

$\text{ And }{D}_3=0\Rightarrow \beta -13=0$

$\begin{array}{l}\Rightarrow \beta =13\\ \therefore \beta -\alpha =13-5=8\end{array}$