fardeen,
1) A point P is denoted by an ordered pair (x,y). Since your line equation is 3x + y = 1, then y = 1 - 3x (I obviously got the signs wrong in my last post, I'm sorry). Since you now know y in terms of x, then P = (x,1-3x).
2) Now that I though a little further into the question, consider this approach: first notice that (AP - AB) is able to assume every value in the real line. So the minimum value of the modulus must be zero. This obviously happens when P is equidistant of both A and B.
So your condition is:$\displaystyle d(P,A) = d(P,B)$
$\displaystyle \sqrt{(x + 2)^2 + (3x + 2)^2} = \sqrt{(x - 3)^2 + (3x - 1)^2}$
Now you square both sides:$\displaystyle (x + 2)^2 + (3x + 2)^2 = (x - 3)^2 + (3x - 1)^2$
This is a quadratic equation. Solve it for x and you will find the value of x for which distances are equal, and because of such the difference is zero.
Hope this helps,