$$D = \left| {\matrix{ 1 & 1 & 1 \cr 1 & 2 & 3 \cr 1 & 3 & \alpha \cr } } \right| = \left| {\matrix{ 1 & 1 & 1 \cr 0 & 1 & 2 \cr 0 & 2 & {\alpha - 1} \cr } } \right|$$

      $$ = \left( {\alpha - 1} \right) - 4 = \left( {\alpha - 5} \right)$$

for infinite solutions $$D = 0 \Rightarrow \alpha = 5$$

$${D_x} = 0 \Rightarrow \left| {\matrix{ 5 & 1 & 1 \cr 9 & 2 & 3 \cr \beta & 3 & 5 \cr } } \right| = 0$$

$$ \Rightarrow \left| {\matrix{ 0 & 0 & 1 \cr { - 1} & { - 1} & 3 \cr {\beta - 15} & { - 2} & 5 \cr } } \right| = 0$$

$$ \Rightarrow 2 + \beta - 15 = 0 \Rightarrow \beta - 13 = 0$$

on $$\beta = 13$$ we get $${D_y} = {D_z} = 0$$

$$\alpha = 5,\beta = 13$$