If $$\frac{\alpha }{{\alpha '}},\,\frac{\beta }{{\beta '}},\,\frac{\gamma }{{\gamma '}}$$   are not all equal, the point of intersection of the lines $$\frac{{x - \alpha '}}{\alpha } = \frac{{y - \beta '}}{\beta } = \frac{{z - \gamma '}}{\gamma }$$      and $$\frac{{x - \alpha }}{{\alpha '}} = \frac{{y - \beta }}{{\beta '}} = \frac{{z - \gamma }}{{\gamma '}}$$      is :

Solution :
From the question, the lines are not parallel.
Any point on the first line is $$\left( {\alpha ' + \alpha r,\,\beta ' + \beta r,\,\gamma ' + \gamma r} \right).$$       It is on the other line if
$$\eqalign{ & \frac{{\alpha ' + \alpha r - \alpha }}{{\alpha '}} = \frac{{\beta ' + \beta r - \beta }}{{\beta '}} = \frac{{\gamma ' + \gamma r - \gamma }}{{\gamma '}} \cr & \Rightarrow 1 + \frac{{\alpha \left( {r - 1} \right)}}{{\alpha '}} = 1 + \frac{{\beta \left( {r - 1} \right)}}{{\beta '}} = 1 + \frac{{\gamma \left( {r - 1} \right)}}{{\gamma '}} \cr & \Rightarrow \frac{{\alpha \left( {r - 1} \right)}}{{\alpha '}} = \frac{{\beta \left( {r - 1} \right)}}{{\beta '}} = \frac{{\gamma \left( {r - 1} \right)}}{{\gamma '}} \cr & \Rightarrow r - 1 = 0\,\,\,\,\,\,\left( {\because \frac{\alpha }{{\alpha '}} = \frac{\beta }{{\beta '}} = \frac{\gamma }{{\gamma '}}{\text{ is not true}}} \right) \cr} $$
$$\therefore $$  the point of intersection is $$\left( {\alpha ' + \alpha .1,\,\beta ' + \beta .1,\,\gamma ' + \gamma .1} \right).$$