Let $$\lambda \in \mathbb{R}$$, $$\vec{a} = \lambda\hat{i} + 2\hat{j} - 3\hat{k}$$, $$\vec{b} = \hat{i} - \lambda\hat{j} + 2\hat{k}$$. If $$\left((\vec{a}+\vec{b}) \times (\vec{a} \times \vec{b})\right) \times (\vec{a}-\vec{b}) = 8\hat{i} - 40\hat{j} - 24\hat{k}$$ then $$\left|\lambda(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})\right|^2$$ is equal to

Given $$\vec{a} = \lambda\hat{i} + 2\hat{j} - 3\hat{k}$$, $$\vec{b} = \hat{i} - \lambda\hat{j} + 2\hat{k}$$, and $$((\vec{a}+\vec{b}) \times (\vec{a} \times \vec{b})) \times (\vec{a}-\vec{b}) = 8\hat{i} - 40\hat{j} - 24\hat{k}$$.

Let $$\vec{p} = \vec{a} + \vec{b}$$, $$\vec{q} = \vec{a} \times \vec{b}$$, $$\vec{r} = \vec{a} - \vec{b}$$.

Using the BAC-CAB rule: $$(\vec{p} \times \vec{q}) \times \vec{r} = \vec{q}(\vec{p} \cdot \vec{r}) - \vec{p}(\vec{q} \cdot \vec{r})$$.

Since $$\vec{q} = \vec{a} \times \vec{b}$$ is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$:

$$\vec{q} \cdot \vec{r} = (\vec{a} \times \vec{b}) \cdot (\vec{a} - \vec{b}) = (\vec{a} \times \vec{b}) \cdot \vec{a} - (\vec{a} \times \vec{b}) \cdot \vec{b} = 0 - 0 = 0$$

Therefore: $$(\vec{p} \times \vec{q}) \times \vec{r} = \vec{q}(\vec{p} \cdot \vec{r})$$.

$$|\vec{a}|^2 = \lambda^2 + 4 + 9 = \lambda^2 + 13$$

$$|\vec{b}|^2 = 1 + \lambda^2 + 4 = \lambda^2 + 5$$

$$\vec{p} \cdot \vec{r} = (\lambda^2 + 13) - (\lambda^2 + 5) = 8$$

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \lambda & 2 & -3 \\ 1 & -\lambda & 2 \end{vmatrix}$$

$$= \hat{i}(2 \cdot 2 - (-3)(-\lambda)) - \hat{j}(\lambda \cdot 2 - (-3) \cdot 1) + \hat{k}(\lambda(-\lambda) - 2 \cdot 1)$$

$$= \hat{i}(4 - 3\lambda) - \hat{j}(2\lambda + 3) + \hat{k}(-\lambda^2 - 2)$$

From Step 2 and 3: $$8(\vec{a} \times \vec{b}) = 8\hat{i} - 40\hat{j} - 24\hat{k}$$

Dividing by 8: $$\vec{a} \times \vec{b} = \hat{i} - 5\hat{j} - 3\hat{k}$$

Comparing the $$\hat{i}$$ component: $$4 - 3\lambda = 1 \implies \lambda = 1$$

Verifying with $$\hat{j}$$: $$-(2(1) + 3) = -5$$ ✓

Verifying with $$\hat{k}$$: $$-(1 + 2) = -3$$ ✓

With $$\lambda = 1$$: $$\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$$, $$\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$$.

$$\vec{a} + \vec{b} = 2\hat{i} + \hat{j} - \hat{k}$$

$$\vec{a} - \vec{b} = 0\hat{i} + 3\hat{j} - 5\hat{k}$$

$$(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 0 & 3 & -5 \end{vmatrix}$$

$$= \hat{i}(1 \cdot (-5) - (-1) \cdot 3) - \hat{j}(2 \cdot (-5) - (-1) \cdot 0) + \hat{k}(2 \cdot 3 - 1 \cdot 0)$$

$$= \hat{i}(-5 + 3) - \hat{j}(-10 - 0) + \hat{k}(6 - 0) = -2\hat{i} + 10\hat{j} + 6\hat{k}$$

Since $$\lambda = 1$$:

$$|\lambda(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|^2 = (-2)^2 + 10^2 + 6^2 = 4 + 100 + 36 = 140$$

The correct answer is Option A: $$140$$.