Show that the vectors a = 1/7(2i + 3j + 6k), b = 1/7(3i - 6j + 2k), and c = 1/7(6i + 2j - 3k) are mutually perpendicular unit vectors.

$$|\vec{a}| = \left| \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}) \right| = \frac{1}{7} \sqrt{2^2 + 3^2 + 6^2}$$ $$|\vec{a}| = \frac{1}{7} \sqrt{4 + 9 + 36} = \frac{1}{7} \sqrt{49} = \frac{1}{7} \times 7 = 1$$

Thus, $\vec{a}$ is a unit vector.

$$|\vec{b}| = \left| \frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k}) \right| = \frac{1}{7} \sqrt{3^2 + (-6)^2 + 2^2}$$ $$|\vec{b}| = \frac{1}{7} \sqrt{9 + 36 + 4} = \frac{1}{7} \sqrt{49} = \frac{1}{7} \times 7 = 1$$

Thus, $\vec{b}$ is a unit vector.

$$|\vec{c}| = \left| \frac{1}{7}(6\hat{i} + 2\hat{j} - 3\hat{k}) \right| = \frac{1}{7} \sqrt{6^2 + 2^2 + (-3)^2}$$ $$|\vec{c}| = \frac{1}{7} \sqrt{36 + 4 + 9} = \frac{1}{7} \sqrt{49} = \frac{1}{7} \times 7 = 1$$

Thus, $\vec{c}$ is a unit vector.
All three vectors are unit vectors.

To show that they are mutually perpendicular, we must show that the dot product of any two distinct vectors is 0. The dot product of $\vec{u} = u_x\hat{i} + u_y\hat{j} + u_z\hat{k}$ and $\vec{v} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}$ is $\vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z$.

$$\vec{a} \cdot \vec{b} = \left( \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}) \right) \cdot \left( \frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k}) \right)$$ $$\vec{a} \cdot \vec{b} = \frac{1}{49} ((2)(3) + (3)(-6) + (6)(2))$$ $$\vec{a} \cdot \vec{b} = \frac{1}{49} (6 - 18 + 12) = \frac{1}{49} (0) = 0$$

Thus, $\vec{a}$ and $\vec{b}$ are perpendicular.

$$\vec{b} \cdot \vec{c} = \left( \frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k}) \right) \cdot \left( \frac{1}{7}(6\hat{i} + 2\hat{j} - 3\hat{k}) \right)$$ $$\vec{b} \cdot \vec{c} = \frac{1}{49} ((3)(6) + (-6)(2) + (2)(-3))$$ $$\vec{b} \cdot \vec{c} = \frac{1}{49} (18 - 12 - 6) = \frac{1}{49} (0) = 0$$

Thus, $\vec{b}$ and $\vec{c}$ are perpendicular.

$$\vec{a} \cdot \vec{c} = \left( \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}) \right) \cdot \left( \frac{1}{7}(6\hat{i} + 2\hat{j} - 3\hat{k}) \right)$$ $$\vec{a} \cdot \vec{c} = \frac{1}{49} ((2)(6) + (3)(2) + (6)(-3))$$ $$\vec{a} \cdot \vec{c} = \frac{1}{49} (12 + 6 - 18) = \frac{1}{49} (0) = 0$$

Thus, $\vec{a}$ and $\vec{c}$ are perpendicular.

Since the dot product of every pair of distinct vectors is 0, the three vectors are mutually perpendicular to each other.

The final answer is $\boxed{\text{Each vector has a magnitude of 1, and the dot product of any two distinct vectors is 0, proving they are unit vectors and mutually perpendicular.}}$