$$\text{Let }\vec a=3\hat i-\hat j+2\hat k,\quad\vec b=\vec a\times(\hat i-2\hat k)\text{ and } \vec c=\vec b\times\hat k.\text{Then the projection of } (\vec c-2\hat j)\text{ on } \vec a \text{ is:}$$

Given $$\mathbf{a} = 3\hat{i} - \hat{j} + 2\hat{k}$$ and $$\mathbf{b} = \mathbf{a} \times (\hat{i} - 2\hat{k})$$:

$$\mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 0 & -2 \end{vmatrix}$$ $$\mathbf{b} = \hat{i}(2 - 0) - \hat{j}(-6 - 2) + \hat{k}(0 + 1)$$ $$\mathbf{b} = 2\hat{i} + 8\hat{j} + \hat{k}$$

Given $$\mathbf{c} = \mathbf{b} \times \hat{k}$$:

$$\mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 8 & 1 \\ 0 & 0 & 1 \end{vmatrix}$$ $$\mathbf{c} = \hat{i}(8 - 0) - \hat{j}(2 - 0) + \hat{k}(0 - 0)$$ $$\mathbf{c} = 8\hat{i} - 2\hat{j}$$

$$\mathbf{c} - 2\hat{j} = (8\hat{i} - 2\hat{j}) - 2\hat{j} = 8\hat{i} - 4\hat{j}$$

The formula for the projection of a vector $$\mathbf{u}$$ onto a vector $$\mathbf{v}$$ is $$\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|}$$.

First, let's find the magnitude of $$\mathbf{a}$$:

$$|\mathbf{a}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$$

$$(\mathbf{c} - 2\hat{j}) \cdot \mathbf{a}$$ = $$(8\hat{i} - 4\hat{j} + 0\hat{k}) \cdot (3\hat{i} - \hat{j} + 2\hat{k})$$= (8)(3) + (-4)(-1) + (0)(2)= 28

$$\text{Projection} = \frac{28}{\sqrt{14}} = \frac{2 \times 14}{\sqrt{14}} = 2\sqrt{14}$$