Let $$\vec{a} = \hat{i} + \alpha\hat{j} + \beta\hat{k}$$, $$\alpha, \beta \in R$$. Let a vector $$\vec{b}$$ be such that the angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\frac{\pi}{4}$$ and $$|\vec{b}|^2 = 6$$. If $$\vec{a} \cdot \vec{b} = 3\sqrt{2}$$, then the value of $$(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2$$ is equal to

$$|\vec{a}|^2 = 1 + \alpha^2 + \beta^2$$. $$|\vec{b}|^2 = 6$$. $$\vec{a} \cdot \vec{b} = 3\sqrt{2}$$.

Angle $$\pi/4$$ between $$\vec{a}$$ and $$\vec{b}$$: $$\cos(\pi/4) = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{3\sqrt{2}}{|\vec{a}|\sqrt{6}}$$.

$$\frac{1}{\sqrt{2}} = \frac{3\sqrt{2}}{|\vec{a}|\sqrt{6}} \Rightarrow |\vec{a}|\sqrt{6} = 3\sqrt{2} \cdot \sqrt{2} = 6 \Rightarrow |\vec{a}| = \sqrt{6}$$.

So $$1 + \alpha^2 + \beta^2 = 6 \Rightarrow \alpha^2 + \beta^2 = 5$$.

$$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 = 36 - 18 = 18$$.

$$(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2 = 5 \times 18 = 90$$.

The answer is Option (1): $$\boxed{90}$$.