The point will be on the line through the origin that has a direction vector equal to the normal vector of the plane: (3, 1, 3).
The line's equation is ...
... (x, y, z) = t·(3, 1, 3)
The parametric equation of that line must satisfy the equation of the plane, so we can find the parameter by ...
... 3(3t) + (t) + 3(3t) = 3
... 9t +t +9t = 3
... 19t = 3
... t = 3/19
The point is (3/19)(3, 1, 3) = (9/19, 3/19, 9/19)