Given,

$$x + y + z = 6$$ ...... (1)

$$2x + 5y + \alpha z = \beta $$ ..... (2)

$$x + 2y + 3z = 14$$ ...... (3)

System of equation have infinite many solutions.

$$\therefore$$ $${\Delta _x} = {\Delta _y} = {\Delta _z} = 0$$ and $$\Delta = 0$$

Now, $$\Delta = \left| {\matrix{ 1 & 1 & 1 \cr 2 & 5 & \alpha \cr 1 & 2 & 3 \cr } } \right| = 0$$

$${C_1} \to {C_1} - {C_3}$$

$${C_2} \to {C_2} - {C_3}$$

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

$$ \Rightarrow - 2 + \alpha + 10 - 2\alpha = 0$$

$$ \Rightarrow 8 - \alpha = 0$$

$$ \Rightarrow \alpha = 8$$

Now, $$x + y + z = 6$$

$$2x + 5y + 8z = \beta $$

$$x + 2y + 3z = 14$$

$$\therefore$$ $${\Delta _x} = \left| {\matrix{ 6 & 1 & 1 \cr \beta & 5 & 8 \cr {14} & 2 & 3 \cr } } \right| = 0$$

$${C_1} \to {C_1} - 6{C_3}$$

$${C_2} \to {C_2} - {C_3}$$

$$ \Rightarrow \left| {\matrix{ 0 & 0 & 1 \cr {\beta - 48} & { - 3} & 8 \cr { - 4} & { - 1} & 3 \cr } } \right| = 0$$

$$ \Rightarrow - \beta + 48 - 12 = 0$$

$$ \Rightarrow \beta = 36$$

$$\therefore$$ $$\alpha + \beta = 8 + 36 = 44$$