Given vectors r1 = 2i + 2j + k and r2 = 6i − 3j + 2k, find the radius and the magnitude.

The magnitude of a vector $\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ is given by $|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$.
For $r_1 = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k}$:

$$|r_1| = \sqrt{2^2 + 2^2 + 1^2}$$ $$|r_1| = \sqrt{4 + 4 + 1}$$ $$|r_1| = \sqrt{9}$$ $$|r_1| = 3$$

The magnitude of $r_1$ is $\boxed{3}$.

For $r_2 = 6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}$:

$$|r_2| = \sqrt{6^2 + (-3)^2 + 2^2}$$ $$|r_2| = \sqrt{36 + 9 + 4}$$ $$|r_2| = \sqrt{49}$$ $$|r_2| = 7$$

The magnitude of $r_2$ is $\boxed{7}$.

First, calculate the difference vector $r_2 - r_1$:

$$r_2 - r_1 = (6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + \mathbf{k})$$ $$r_2 - r_1 = (6-2)\mathbf{i} + (-3-2)\mathbf{j} + (2-1)\mathbf{k}$$ $$r_2 - r_1 = 4\mathbf{i} - 5\mathbf{j} + \mathbf{k}$$

Now, find the magnitude of this difference vector:

$$|r_2 - r_1| = \sqrt{4^2 + (-5)^2 + 1^2}$$ $$|r_2 - r_1| = \sqrt{16 + 25 + 1}$$ $$|r_2 - r_1| = \sqrt{42}$$

The "Radius" (distance between the points) is $\boxed{\sqrt{42}}$.