Solve the Problem $x ^ { 2 } + y ^ { 2 } + 2 z = 4$

Multiple Choice

Solve the problem.
-Find the point on the curve of intersection of the paraboloid $x ^ { 2 } + y ^ { 2 } + 2 z = 4$ and the plane $x - y + 2 z = 0$ that is closest to the origin.

A) (1, 1, 1)
B) (1, -1, 1)
C) (-1, 1, 1)
D) (1, 1, -1)