Given system of equations

$$\alpha x + y + z = 5$$

$$x + 2y + 3z = 4$$, has infinite solution

$$x + 3y + 5z = \beta $$

$$\therefore$$ $$\Delta = \left| {\matrix{ \alpha & 1 & 1 \cr 1 & 2 & 3 \cr 1 & 3 & 5 \cr } } \right| = 0 \Rightarrow \alpha (1) - 1(2) + 1(1) = 0$$

$$ \Rightarrow \alpha = 1$$

and $${\Delta _1} = \left| {\matrix{ 5 & 1 & 1 \cr 4 & 2 & 3 \cr \beta & 3 & 5 \cr } } \right| = 0$$

$$ \Rightarrow 5(1) - 1(20 - 3\beta ) + 1(12 - 2\beta ) = 0$$

$$ \Rightarrow \beta = 3$$

and $${\Delta _2} = \left| {\matrix{ 1 & 5 & 1 \cr 1 & 4 & 3 \cr 1 & \beta & 5 \cr } } \right| = 0 \Rightarrow (20 - 3\beta ) - 5(2) + 1(\beta - 4) = 0$$

$$ \Rightarrow - 2\beta + 6 = 0$$

$$ \Rightarrow \beta = 3$$

Similarly,

$$ \Rightarrow {\Delta _3} = \left| {\matrix{ 1 & 1 & 5 \cr 1 & 2 & 4 \cr 1 & 3 & \beta \cr } } \right| = 0 \Rightarrow \beta = 3$$

$$\therefore$$ ($$\alpha$$, $$\beta$$) = (1, 3)