Let $(x_1,y_1),(x_2,y_2),(x_2,y_2)$ are the vertices of the triangle T.

I want to show that the line $L(\alpha_3)$ defined by $$x=(1-\alpha_3-\alpha)x_1+\alpha x_2+\alpha_3 x_3$$ $$y=(1-\alpha_3-\alpha)y_1+\alpha y_2+\alpha_3 y_3$$ where $(\alpha_3 \in (0,1)~and~\alpha\in R)$ is parallel to the edege of triangle T joining $(x_1,y_1) ~and~(x_2,y_2)$, and I need to find the distance between the line $L(\alpha_3)$ and the the line through $(x_1,y_1),(x_2,y_2)$.

Thanks

I am not sure how to approach this, since it is not clear what results can be taken for granted. Note that for the purposes of the problem, $\alpha_3$ is a fixed number.
The expressions for $x$ and $y$ can be rewritten as $$x=(1-\alpha)x_1+\alpha x_2+\alpha_3(x_3-x_1),$$ $$y=(1-\alpha)y_1+\alpha y_2+\alpha_3 (y_3-y_1).$$ Note that $x=(1-\alpha)x_1+\alpha x_2)$, $y=(1-\alpha)x_1+\alpha y_2$ is the parametric equation of the line through $(x_1,y_1)$ and $(x_2,y_2)$.
So the expressions you were given take an arbitrary point on the line through $(x_1,y_1)$ and add to it the fixed vector $\left(\alpha_3(x_3-x_1), \alpha_3 (y_3-y_1)\right)$. That proves the parallelism.
Another way to prove the parallelism is as follows. Let $\alpha=0$. That gives you a particular point $(x_0,y_0)$. Now take a general $\alpha\ne 0$, and call the point you get $(x_\alpha,y_\alpha)$. (These are just the $x$ and $y$ in the formula of the post.) Show that the line through $(x_0,y_0)$ and $(x_\alpha,y_\alpha)$ is parallel to the line through $(x_1,y_1)$ and $(x_2,y_2)$, perhaps by comparing slopes, or by noting that one difference vector is a constant multiple of the other.
As to the distance part of your question, I do not know what to suggest. The "right" thing to do depends on what machinery you currently have available. Perhaps if I get information about that machinery, I can make a suggestion. If you know how to find the distance from a point to a line, you can pick a simple point on the line, say by choosing $1-\alpha=\alpha_3$, and calculate that way.