Let $$\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$$, $$\vec{b} = ((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}) \times \hat{i}$$. Then the square of the projection of $$\vec{a}$$ on $$\vec{b}$$ is :

Given $$\vec{a}=2\,\hat{i}+\,\hat{j}-\,\hat{k}$$ and $$\vec{b}=((\vec{a}\times(\hat{i}+\hat{j}))\times\hat{i})\times\hat{i}$$. We first evaluate the cross products one by one.

Step 1: Compute $$\vec{c}=\vec{a}\times(\hat{i}+\hat{j})$$.
Write $$\hat{i}+\hat{j}=(1,1,0)$$ and $$\vec{a}=(2,1,-1)$$.
Using the determinant formula for $$\vec{u}\times\vec{v}$$ :

$$\vec{c}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 1 & -1\\ 1 & 1 & 0 \end{vmatrix} =\hat{i}(1\cdot0-(-1)\cdot1)-\hat{j}(2\cdot0-(-1)\cdot1)+\hat{k}(2\cdot1-1\cdot1)$$

$$\Rightarrow\;\vec{c}=1\,\hat{i}-1\,\hat{j}+1\,\hat{k}=(1,-1,1)$$.

Step 2: Compute $$\vec{d}=\vec{c}\times\hat{i}$$.
Take $$\hat{i}=(1,0,0)$$:

$$\vec{d}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & -1 & 1\\ 1 & 0 & 0 \end{vmatrix} =\hat{i}((-1)\cdot0-1\cdot0)-\hat{j}(1\cdot0-1\cdot1)+\hat{k}(1\cdot0-(-1)\cdot1)$$

$$\Rightarrow\;\vec{d}=0\,\hat{i}+1\,\hat{j}+1\,\hat{k}=(0,1,1)$$.

Step 3: Compute $$\vec{b}=\vec{d}\times\hat{i}$$.
Again cross with $$\hat{i}=(1,0,0)$$ :

$$\vec{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 0 & 1 & 1\\ 1 & 0 & 0 \end{vmatrix} =\hat{i}(1\cdot0-1\cdot0)-\hat{j}(0\cdot0-1\cdot1)+\hat{k}(0\cdot0-1\cdot1)$$

$$\Rightarrow\;\vec{b}=0\,\hat{i}+1\,\hat{j}-1\,\hat{k}=(0,1,-1)$$.

Step 4: Square of the projection of $$\vec{a}$$ on $$\vec{b}$$.
Projection length formula: the component of $$\vec{a}$$ along $$\vec{b}$$ is $$\dfrac{\vec{a}\cdot\vec{b}}{\lvert\vec{b}\rvert}$$. Hence

$$\text{(projection)}^{2}=\dfrac{(\vec{a}\cdot\vec{b})^{2}}{\lvert\vec{b}\rvert^{2}}$$.

Dot product: $$\vec{a}\cdot\vec{b}=(2,1,-1)\cdot(0,1,-1)=0+1+1=2$$.
Magnitude squared of $$\vec{b}$$: $$\lvert\vec{b}\rvert^{2}=0^{2}+1^{2}+(-1)^{2}=2$$.

Therefore $$\text{(projection)}^{2}=\dfrac{2^{2}}{2}=2$$.

Hence the square of the projection of $$\vec{a}$$ on $$\vec{b}$$ is $$2$$, which matches Option C.