Let $$\vec{a} = \hat{i} + \hat{j} + \hat{k}$$, $$\vec{c} = \hat{j} - \hat{k}$$ and a vector $$\vec{b}$$ be such that $$\vec{a} \times \vec{b} = \vec{c}$$ and $$\vec{a} \cdot \vec{b} = 3$$. Then $$|\vec{b}|$$ equals:

We have the three-dimensional vectors

$$\vec a = \hat i + \hat j + \hat k,$$

$$\vec c = \hat j - \hat k,$$

and an unknown vector $$\vec b = b_x\hat i + b_y\hat j + b_z\hat k$$ which must satisfy two simultaneous conditions:

$$\vec a \times \vec b = \vec c \qquad\text{and}\qquad \vec a \cdot \vec b = 3.$$

We attack the cross-product equation first. Using the determinant form of the vector product

$$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\[4pt] 1 & 1 & 1 \\[4pt] b_x & b_y & b_z \end{vmatrix},$$

we obtain, expanding along the first row,

$$\vec a \times \vec b = (1\cdot b_z - 1\cdot b_y)\,\hat i - (1\cdot b_z - 1\cdot b_x)\,\hat j + (1\cdot b_y - 1\cdot b_x)\,\hat k.$$

Simplifying each component gives

$$\vec a \times \vec b = (\,b_z - b_y\,)\,\hat i + (\,b_x - b_z\,)\,\hat j + (\,b_y - b_x\,)\,\hat k.$$

This must equal $$\vec c = 0\,\hat i + 1\,\hat j - 1\,\hat k,$$ so by equating corresponding components we arrive at the three scalar equations

$$$b_z - b_y = 0, \tag{1}$$$

$$$b_x - b_z = 1, \tag{2}$$$

$$$b_y - b_x = -1. \tag{3}$$$

From equation (1) we have

$$$b_z = b_y. \tag{4}$$$

Substituting (4) into equation (2) yields

$$$b_x = 1 + b_z. \tag{5}$$$

Equation (3) can be rearranged to

$$$b_y = b_x - 1. \tag{6}$$$

Now insert (6) into (4) to keep everything in terms of $$b_x$$:

$$$b_z = b_y = b_x - 1. \tag{7}$$$

We next impose the dot-product condition. The scalar (dot) product formula is

$$\vec a \cdot \vec b = a_x b_x + a_y b_y + a_z b_z.$$

Because $$\vec a = (1,1,1),$$ this becomes simply the sum of the components of $$\vec b$$:

$$$\vec a \cdot \vec b = b_x + b_y + b_z = 3. \tag{8}$$$

Using (6) and (7) to replace $$b_y$$ and $$b_z$$ in (8) we get

$$b_x + (b_x - 1) + (b_x - 1) = 3.$$

Simplifying the left side gives

$$3b_x - 2 = 3,$$

so

$$3b_x = 5$$

and therefore

$$$b_x = \frac{5}{3}. \tag{9}$$$

Using (6) we find

$$b_y = b_x - 1 = \frac{5}{3} - 1 = \frac{2}{3},$$

and from (7)

$$b_z = \frac{2}{3}.$$

Thus the explicit vector is

$$\vec b = \frac{5}{3}\,\hat i + \frac{2}{3}\,\hat j + \frac{2}{3}\,\hat k.$$

Finally we compute its magnitude. The formula for the magnitude of a vector $$\vec v = (v_x, v_y, v_z)$$ is

$$|\vec v| = \sqrt{v_x^2 + v_y^2 + v_z^2}.$$

Applying this to $$\vec b$$ we have

$$|\vec b| = \sqrt{\left(\frac{5}{3}\right)^2 + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^2} = \sqrt{\frac{25}{9} + \frac{4}{9} + \frac{4}{9}} = \sqrt{\frac{33}{9}} = \sqrt{\frac{11}{3}}.$$

Hence, the correct answer is Option C.