Let $$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$$, $$\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$$ and $$\vec{c} = \hat{i} + \hat{j} - 2\hat{k}$$ be three vectors. A vector of the type $$\vec{b} + \lambda\vec{c}$$ for some scalar $$\lambda$$, whose projection on $$\vec{a}$$ is of magnitude $$\sqrt{\frac{2}{3}}$$ is :

We are given three vectors: $$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$$, $$\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$$, and $$\vec{c} = \hat{i} + \hat{j} - 2\hat{k}$$. We need to find a vector of the form $$\vec{b} + \lambda\vec{c}$$ for some scalar $$\lambda$$ such that the magnitude of its projection on $$\vec{a}$$ is $$\sqrt{\frac{2}{3}}$$.

The projection of a vector $$\vec{u}$$ onto $$\vec{a}$$ is given by $$\frac{\vec{u} \cdot \vec{a}}{|\vec{a}|}$$, and its magnitude is $$\left| \frac{\vec{u} \cdot \vec{a}}{|\vec{a}|} \right|$$. Here, $$\vec{u} = \vec{b} + \lambda\vec{c}$$, so we set up the equation:

$$\left| \frac{(\vec{b} + \lambda\vec{c}) \cdot \vec{a}}{|\vec{a}|} \right| = \sqrt{\frac{2}{3}}$$

Expanding the dot product:

$$\left| \frac{\vec{b} \cdot \vec{a} + \lambda (\vec{c} \cdot \vec{a})}{|\vec{a}|} \right| = \sqrt{\frac{2}{3}}$$

First, compute the dot products and the magnitude of $$\vec{a}$$.

Calculate $$\vec{a} \cdot \vec{b}$$:

$$\vec{a} \cdot \vec{b} = (2)(1) + (-1)(2) + (1)(-1) = 2 - 2 - 1 = -1$$

Calculate $$\vec{a} \cdot \vec{c}$$:

$$\vec{a} \cdot \vec{c} = (2)(1) + (-1)(1) + (1)(-2) = 2 - 1 - 2 = -1$$

Calculate $$|\vec{a}|$$:

$$|\vec{a}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$$

Substitute these values into the equation:

$$\left| \frac{-1 + \lambda (-1)}{\sqrt{6}} \right| = \sqrt{\frac{2}{3}}$$

Simplify the numerator:

$$\left| \frac{-1 - \lambda}{\sqrt{6}} \right| = \sqrt{\frac{2}{3}}$$

Since the absolute value, we can write:

$$\frac{| -1 - \lambda |}{\sqrt{6}} = \sqrt{\frac{2}{3}}$$

Note that $$| -1 - \lambda | = | -(1 + \lambda) | = |1 + \lambda|$$, so:

$$\frac{|1 + \lambda|}{\sqrt{6}} = \sqrt{\frac{2}{3}}$$

Multiply both sides by $$\sqrt{6}$$:

$$|1 + \lambda| = \sqrt{6} \cdot \sqrt{\frac{2}{3}}$$

Simplify the right side:

$$\sqrt{6} \cdot \sqrt{\frac{2}{3}} = \sqrt{6 \cdot \frac{2}{3}} = \sqrt{4} = 2$$

So:

$$|1 + \lambda| = 2$$

This gives two cases:

Case 1: $$1 + \lambda = 2$$, so $$\lambda = 2 - 1 = 1$$

Case 2: $$1 + \lambda = -2$$, so $$\lambda = -2 - 1 = -3$$

Now, form the vector $$\vec{b} + \lambda\vec{c}$$ for each $$\lambda$$.

For $$\lambda = 1$$:

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

For $$\lambda = -3$$:

$$\vec{b} + \lambda\vec{c} = (\hat{i} + 2\hat{j} - \hat{k}) + (-3) \cdot (\hat{i} + \hat{j} - 2\hat{k}) = \hat{i} + 2\hat{j} - \hat{k} - 3\hat{i} - 3\hat{j} + 6\hat{k} = (1 - 3)\hat{i} + (2 - 3)\hat{j} + (-1 + 6)\hat{k} = -2\hat{i} - \hat{j} + 5\hat{k}$$

Now, compare these vectors with the given options:

A. $$2\hat{i} + \hat{j} + 5\hat{k}$$

B. $$2\hat{i} + 3\hat{j} - 3\hat{k}$$

C. $$2\hat{i} - \hat{j} + 5\hat{k}$$

D. $$2\hat{i} + 3\hat{j} + 3\hat{k}$$

The vector for $$\lambda = 1$$ is $$2\hat{i} + 3\hat{j} - 3\hat{k}$$, which matches option B.

The vector for $$\lambda = -3$$ is $$-2\hat{i} - \hat{j} + 5\hat{k}$$, which does not match any option.

Verify that the vector for $$\lambda = 1$$ satisfies the projection magnitude condition:

Vector: $$\vec{u} = 2\hat{i} + 3\hat{j} - 3\hat{k}$$

Dot product with $$\vec{a}$$: $$\vec{u} \cdot \vec{a} = (2)(2) + (3)(-1) + (-3)(1) = 4 - 3 - 3 = -2$$

Magnitude of $$\vec{a}$$: $$|\vec{a}| = \sqrt{6}$$

Projection magnitude: $$\left| \frac{-2}{\sqrt{6}} \right| = \frac{2}{\sqrt{6}} = \sqrt{\frac{4}{6}} = \sqrt{\frac{2}{3}}$$, which matches.

Although the vector for $$\lambda = -3$$ also satisfies the condition, it is not listed in the options. Only option B matches one of the solutions and is present in the choices.

Hence, the correct answer is Option B.