Let $$\vec{a}, \vec{b}, \vec{c}$$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $$\theta$$, with the vector $$\vec{a} + \vec{b} + \vec{c}$$. Then $$36\cos^2 2\theta$$ is equal to ___.

Let $$|\vec{a}| = |\vec{b}| = |\vec{c}| = k$$. Since $$\vec{a}, \vec{b}, \vec{c}$$ are mutually perpendicular, all dot products between distinct vectors are zero. Let $$\vec{s} = \vec{a} + \vec{b} + \vec{c}$$. Then:

$$|\vec{s}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 = 3k^2 \implies |\vec{s}| = k\sqrt{3}$$

The angle $$\theta$$ that $$\vec{a}$$ makes with $$\vec{s}$$ satisfies:

$$\cos\theta = \frac{\vec{a} \cdot \vec{s}}{|\vec{a}||\vec{s}|} = \frac{\vec{a}\cdot(\vec{a}+\vec{b}+\vec{c})}{k \cdot k\sqrt{3}} = \frac{k^2}{k^2\sqrt{3}} = \frac{1}{\sqrt{3}}$$

By symmetry, $$\vec{b}$$ and $$\vec{c}$$ make the same angle with $$\vec{s}$$, confirming they are equally inclined. Now:

$$\cos 2\theta = 2\cos^2\theta - 1 = \frac{2}{3} - 1 = -\frac{1}{3}$$

$$36\cos^2 2\theta = 36 \cdot \frac{1}{9} = 4$$