Let $$\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$$ and $$\vec{b} = 2\hat{i} - 3\hat{j} + 5\hat{k}$$. If $$\vec{r} \times \vec{a} = \vec{b} \times \vec{r}$$, $$\vec{r} \cdot (\alpha\hat{i} + 2\hat{j} + \hat{k}) = 3$$ and $$\vec{r} \cdot (2\hat{i} + 5\hat{j} - \alpha\hat{k}) = -1$$, $$\alpha \in R$$, then the value of $$\alpha + |\vec{r}|^2$$ is equal to:

We have $$\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$$ and $$\vec{b} = 2\hat{i} - 3\hat{j} + 5\hat{k}$$. The condition $$\vec{r} \times \vec{a} = \vec{b} \times \vec{r}$$ can be rewritten as $$\vec{r} \times \vec{a} + \vec{r} \times \vec{b} = \vec{0}$$, i.e., $$\vec{r} \times (\vec{a} + \vec{b}) = \vec{0}$$.

This means $$\vec{r}$$ is parallel to $$\vec{a} + \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$$. So $$\vec{r} = t(3\hat{i} - \hat{j} + 2\hat{k})$$ for some scalar $$t$$.

From the first dot product condition: $$\vec{r} \cdot (\alpha\hat{i} + 2\hat{j} + \hat{k}) = 3$$, so $$t(3\alpha - 2 + 2) = 3$$, giving $$3\alpha t = 3$$, hence $$t = \frac{1}{\alpha}$$.

From the second dot product condition: $$\vec{r} \cdot (2\hat{i} + 5\hat{j} - \alpha\hat{k}) = -1$$, so $$t(6 - 5 - 2\alpha) = -1$$, giving $$t(1 - 2\alpha) = -1$$.

Substituting $$t = \frac{1}{\alpha}$$: $$\frac{1 - 2\alpha}{\alpha} = -1$$, so $$1 - 2\alpha = -\alpha$$, giving $$\alpha = 1$$.

Then $$t = 1$$ and $$\vec{r} = 3\hat{i} - \hat{j} + 2\hat{k}$$, so $$|\vec{r}|^2 = 9 + 1 + 4 = 14$$.

Therefore, $$\alpha + |\vec{r}|^2 = 1 + 14 = 15$$.