I am pretty sure I have the answer but I am not quite sure how. Here is what I have done so far.
$$ a(t)=\langle 1 , 2, -2\rangle \\ v(t)= \int a(t) = \langle t+c_1 , 2t+c_2, -2t+c_3\rangle \\ $$
I just guessed and check and found that if I set: $$c_1 = 4\\ c_2 = 2\\ c_3 = 4$$
then $$ |v(t)| = \sqrt{(t+4)^2 + (2t+2)^2 + (-2t+4)^2} \\ =3 \sqrt{t^2+4} $$ and if you evaluate that at time 0 you get $$3*2 = 6$$.
My problem is that I just stumbled into this and for one I am not 100% sure this is the solution but also if given a harder version of this problem I don't know the proper way to solve this.
No
Integrate the acceleration to get velocity;
$(t+b,2t+c,-2t+d)$ (which you found but you mess up afterwards)
You don’t have to guess anything since you are given the speed at time zero. So:
$6=\sqrt{(t+b)^2+(2t+c)^2+(-2t+d)^2}$ and set $t=0$ to get:
$6=\sqrt{b^2+c^2+d^2}$ so for all values of $(b,c,d) that satisfy this condition. (No guess work and you have already found one). Then integrate one more time to get position vector and realize that there are no new constants since the particle goes through the origin.