Let the straight line L be y=k(x- 1) (obtained from oblique points).

The line l and 3x+y-6=0 form an equation group, and x = 6+k/3+k y = 3k/3+k.

The straight line L and 3x+y+3=0 form a set of equations, and x = k-3/3+k y =-6k/3+k.

(x2-x1) 2+(y2-y1) 2 = (9/3+k) 2+(9k/3+k) 2 = 3 (formula between two points).

Solution: k =-2/3

So the straight line L is y =-2/3 (x- 1), which is simplified as 2x+3y-2=0.