For $$\lambda > 0$$, let $$\theta$$ be the angle between the vectors $$\vec{a} = \hat{i} + \lambda\hat{j} - 3\hat{k}$$ and $$\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$$. If the vectors $$\vec{a} + \vec{b}$$ and $$\vec{a} - \vec{b}$$ are mutually perpendicular, then the value of $$(14 \cos \theta)^2$$ is equal to

Since $$(\vec{a} + \vec{b}) \perp (\vec{a} - \vec{b})$$, their dot product is zero: $$|\vec{a}|^2 = |\vec{b}|^2$$.

Equating magnitudes: $$1^2 + \lambda^2 + (-3)^2 = 3^2 + (-1)^2 + 2^2 \implies \lambda^2 + 10 = 14 \implies \lambda = 2$$ (since $$\lambda > 0$$).

Dot product $$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-1) + (-3)(2) = -5$$.

$$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{-5}{\sqrt{14}\sqrt{14}} = -\frac{5}{14}$$.

Evaluating $$(14 \cos \theta)^2 = (-5)^2 = 25$$.