Let $$\vec{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$$, $$\vec{b} = \hat{i} + \hat{k}$$ and $$\vec{c} = \hat{i} + 2\hat{j} - 3\hat{k}$$ be three given vectors. If $$\vec{r}$$ is a vector such that $$\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$$ and $$\vec{r} \cdot \vec{b} = 0$$, then $$|\vec{r}|$$ is equal to:

Given $$\vec{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$$, $$\vec{b} = \hat{i} + \hat{k}$$ and $$\vec{c} = \hat{i} + 2\hat{j} - 3\hat{k}$$, we seek $$|\vec{r}|$$ such that $$\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$$ and $$\vec{r} \cdot \vec{b} = 0$$. From the cross product condition we get $$(\vec{r} - \vec{c}) \times \vec{a} = \vec{0}$$, which implies that $$\vec{r} - \vec{c}$$ is parallel to $$\vec{a}$$, so $$\vec{r} = \vec{c} + \lambda\vec{a}$$ for some scalar $$\lambda$$.

Expressing $$\vec{r}$$ in component form gives $$\vec{r} = (1 + 2\lambda)\hat{i} + (2 - 7\lambda)\hat{j} + (-3 + 5\lambda)\hat{k}$$. The condition $$\vec{r} \cdot \vec{b} = 0$$ yields $$(1 + 2\lambda)(1) + (2 - 7\lambda)(0) + (-3 + 5\lambda)(1) = 0$$, which simplifies to $$1 + 2\lambda - 3 + 5\lambda = 0$$ or $$7\lambda - 2 = 0$$, giving $$\lambda = \frac{2}{7}$$.

Substituting back, $$\vec{r} = \left(1 + \frac{4}{7}\right)\hat{i} + (2 - 2)\hat{j} + \left(-3 + \frac{10}{7}\right)\hat{k} = \frac{11}{7}\hat{i} + 0\hat{j} - \frac{11}{7}\hat{k}$$ and hence $$|\vec{r}| = \sqrt{\left(\frac{11}{7}\right)^2 + 0^2 + \left(-\frac{11}{7}\right)^2} = \sqrt{\frac{242}{49}} = \frac{11}{7}\sqrt{2}$$.

The correct answer is Option (1): $$\frac{11}{7}\sqrt{2}$$.