Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$. We need to evaluate the Left Hand Side (LHS) expression $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b})$.

First, we determine the component forms of the vectors involved using the rules for vector addition and subtraction provided in the context. We have $\vec{a} - \vec{b} = (a_1 - b_1)\hat{i} + (a_2 - b_2)\hat{j} + (a_3 - b_3)\hat{k}$ and $\vec{a} + \vec{b} = (a_1 + b_1)\hat{i} + (a_2 + b_2)\hat{j} + (a_3 + b_3)\hat{k}$.

Next, we compute the cross product using the determinant form. The expression becomes:

$$(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1-b_1 & a_2-b_2 & a_3-b_3 \\ a_1+b_1 & a_2+b_2 & a_3+b_3 \end{vmatrix}$$

Expanding this determinant, we get: $= \hat{i}[(a_2-b_2)(a_3+b_3) - (a_3-b_3)(a_2+b_2)] - \hat{j}[(a_1-b_1)(a_3+b_3) - (a_3-b_3)(a_1+b_1)] + \hat{k}[(a_1-b_1)(a_2+b_2) - (a_2-b_2)(a_1+b_1)]$.

Simplifying the coefficients algebraically, for $\hat{i}$ we get $2(a_2b_3 - a_3b_2)$. Similarly, for $\hat{j}$ we get $-2(a_1b_3 - a_3b_1)$ and for $\hat{k}$ we get $2(a_1b_2 - a_2b_1)$. Substituting these back, the expression becomes:

$$= 2(a_2b_3 - a_3b_2)\hat{i} - 2(a_1b_3 - a_3b_1)\hat{j} + 2(a_1b_2 - a_2b_1)\hat{k}$$

$= 2 [(a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}]$.

From the context, the term inside the bracket is exactly the expansion of $\vec{a} \times \vec{b}$. Therefore, LHS = $2(\vec{a} \times \vec{b})$, which equals the Right Hand Side (RHS). Hence proved.