So I used the trig identity (y^2 + z^2 = 1) on my y and z component. So I concluded that the cylinder y^2 + z^2 = 4 satisfies the question.
I also concluded that the plane x + y = 3 satisfies the question using the original vector I was given.
I then attempted to graph it, and I am under the impression that it is a sinusoid on the plane x+y = 3. I don't understand how it lies on (y^2 + z^2 = 1), so I'm doubting whether I approached this problem correctly. I thought that for the answer to be E (for the vector to lie on multiple surfaces that the shape would have to be a different shape than a sinusoid.
Thank you.
Your conclusion that it lies on ii. and iii. is correct. This can be seen by plugging your vector components into each equation:
$y^2+z^2 = (2cos(t))^2+(2sin(t))^2 = 4(sin(t)^2+cos(t)^2) = 4$ and
$x+y = 3-2cos(t)+2cos(t) = 3$.