First, notice that parallel to y=x-3 are all functions with slope 1 (all functions y=x+b where b is an arbitrary real number).

Second, most generally, normal to the linear function y=ax+b is parallel to y=(-1/a)x

Now, tangent to y=3x^2-2x-1 at x=A is parallel to y'=(6A-2)x. Normal to y' is parallel to y''=[-1/(6A-2)]x.

So, if we want y'' to be parallel to y=x-3 we obtain a condition:

[-1/(6A-2)]=1 (solution: A=1/6)

Thus, our normal is given by y''=x+b where b is such that at x=A y'' and y intersect, so:

3A^2-2A-1 = A+b (solution: b=-17/12)

All in all, function we were seeking is given by y''=x-17/12

Hope this will help!