Excellent algebraic have been given (I voted for them). Intuition can here be obtained through visual proofs.
The equation $x+y+z=1$ defines a plane. $x^2+y^2+z^2=1/3$ defines a sphere. The following visualization depicts the plane in blue, the sphere in red.

From that, you can imagine that the question could be rephrased as (in a mundane way): for any point in the plane, its distance from the $(0,0,0)$ origin is higher than $1/\sqrt{3}$? So the sphere should remain "below" the plane. Except when they meet. The symmetry of the problem tells you that the tangency point has equal coordinates $(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})$, which is one of the motivations behind the equation $(3x-1)^2+(3y-1)^2+(3z-1)^2$.
Would the sphere be bigger (a radius higher than $1/\sqrt{3}$), it would intersect the plane in more than a single tangency point.
All in all, this resorts to finding the distance of the plane to the origin, which is exactly where the sphere and the plane meet. So what you are looking at is the distance of the plane to the origin.
If the plane is given by $ax+by+cz+d$, the signed distance of a point $(x_0,y_0,z_0)$ to it is (Point-Plane Distance):
$$D = \frac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}$$
which in your case gives exactly $1/\sqrt{3}$. Any point in the plane is farther to $(0,0,0)$ than $|D|$.