Let $$\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$$ and let $$\vec{b}$$ be a vector such that $$\vec{a} \times \vec{b} = 2\hat{i} - \hat{k}$$ and $$\vec{a} \cdot \vec{b} = 3$$. Then the projection of $$\vec{b}$$ on the vector $$\vec{a} - \vec{b}$$ is:

We are given $$\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$$, $$\vec{a} \times \vec{b} = 2\hat{i} - \hat{k}$$, and $$\vec{a} \cdot \vec{b} = 3$$. To determine $$\vec{b}$$, let $$\vec{b}=(b_1,b_2,b_3)$$. The cross product in component form becomes

$$ (-b_3 - 2b_2,\;2b_1 - b_3,\;b_1 + b_2) \;=\;(2,0,-1) $$

Equating components gives

$$-b_3 - 2b_2 = 2\quad(i),\qquad 2b_1 - b_3 = 0\quad(ii),\qquad b_1 + b_2 = -1\quad(iii).$$

From (ii) we have $$b_3 = 2b_1$$, and from (iii) $$b_2 = -1 - b_1$$. Substituting these into the dot-product condition $$\vec{a}\cdot\vec{b}=3$$ yields

$$b_1 - b_2 + 2b_3 = 3 \;\implies\; b_1 + 1 + b_1 + 4b_1 = 3 \;\implies\; 6b_1 = 2 \;\implies\; b_1 = \tfrac{1}{3}.$$

Thus

$$\vec{b}=\Bigl(\tfrac{1}{3},\,-\tfrac{4}{3},\,\tfrac{2}{3}\Bigr).$$

Subtracting this from $$\vec{a}$$ gives

$$\vec{a}-\vec{b}=\Bigl(\tfrac{2}{3},\;\tfrac{1}{3},\;\tfrac{4}{3}\Bigr).$$

Next, the scalar product of $$\vec{b}$$ with $$\vec{a}-\vec{b}$$ is

$$ \frac{1}{3}\cdot\frac{2}{3} \;+\;\bigl(-\tfrac{4}{3}\bigr)\cdot\tfrac{1}{3} \;+\;\tfrac{2}{3}\cdot\tfrac{4}{3} \;=\;\frac{2}{9}-\frac{4}{9}+\frac{8}{9} \;=\;\frac{6}{9} \;=\;\frac{2}{3}. $$

The magnitude of $$\vec{a}-\vec{b}$$ is

$$ \sqrt{\frac{4}{9}+\frac{1}{9}+\frac{16}{9}} \;=\;\sqrt{\frac{21}{9}} \;=\;\frac{\sqrt{21}}{3}. $$

Therefore, the projection of $$\vec{b}$$ onto $$\vec{a}-\vec{b}$$ is

$$ \frac{\vec{b}\cdot(\vec{a}-\vec{b})}{\bigl|\vec{a}-\vec{b}\bigr|} \;=\;\frac{\tfrac{2}{3}}{\tfrac{\sqrt{21}}{3}} \;=\;\frac{2}{\sqrt{21}}. $$

The answer is Option A: $$\dfrac{2}{\sqrt{21}}$$.