For the following two (02) items:
Suppose $S$ is the sphere with the smallest radius that passes through the points $A(1, 0, 0)$, $B(0, 1, 0)$ and $C(0, 0, 1)$.

On which one of the following planes does the centre of $S$ lie?

The sphere $S$ passes through the points $A(1, 0, 0)$, $B(0, 1, 0)$, and $C(0, 0, 1)$.

The equation of the plane containing these three points is given by the intercept form:

$\dfrac{x}{1} + \dfrac{y}{1} + \dfrac{z}{1} = 1 \Rightarrow x + y + z - 1 = 0$

Any sphere passing through $A, B$, and $C$ will intersect this plane in a circle, which is the circumcircle of $\triangle ABC$.

For the sphere to have the smallest possible radius, this circumcircle must be a great circle of the sphere. This means the centre of the sphere must lie exactly on the plane containing the points $A, B$, and $C$.

Thus, the centre of the sphere $S$ lies on the plane $x + y + z - 1 = 0$.

Alternatively, $\triangle ABC$ is an equilateral triangle. Its circumcentre is the same as its centroid, which is $\left(\dfrac{1+0+0}{3}, \dfrac{0+1+0}{3}, \dfrac{0+0+1}{3}\right) = \left(\dfrac{1}{3}, \dfrac{1}{3}, \dfrac{1}{3}\right)$.

Substituting $\left(\dfrac{1}{3}, \dfrac{1}{3}, \dfrac{1}{3}\right)$ into the given options, only $x + y + z - 1 = 0$ is satisfied.

Answer: $x + y + z - 1 = 0$