Let $$\vec{a} = 3\hat{i} + \hat{j}$$ and $$\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$$. Let $$\vec{c}$$ be a vector satisfying $$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda\vec{c}$$. If $$\vec{b}$$ and $$\vec{c}$$ are non-parallel, then the value of $$\lambda$$ is

We have $$\vec{a} = 3\hat{i} + \hat{j}$$, $$\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$$, and the relation $$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda\vec{c}$$.

We apply the BAC-CAB identity for the vector triple product: $$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$$.

We compute $$\vec{a} \cdot \vec{b} = 3(1) + 1(2) + 0(1) = 5$$. So the equation becomes:

$$\vec{b}(\vec{a} \cdot \vec{c}) - 5\vec{c} = \vec{b} + \lambda\vec{c}$$

Rearranging: $$(\vec{a} \cdot \vec{c} - 1)\vec{b} = (5 + \lambda)\vec{c}$$.

Since $$\vec{b}$$ and $$\vec{c}$$ are non-parallel, neither can be expressed as a scalar multiple of the other. The only way the equation $$(\vec{a} \cdot \vec{c} - 1)\vec{b} = (5 + \lambda)\vec{c}$$ can hold with $$\vec{b}$$ and $$\vec{c}$$ non-parallel is if both coefficients are zero. That is, $$\vec{a} \cdot \vec{c} - 1 = 0$$ and $$5 + \lambda = 0$$.

From the second condition, $$\lambda = -5$$.

Hence, the correct answer is Option A.