Let $$\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$$ and $$\vec{b} = \hat{i} - \hat{j} + \hat{k}$$, for some real $$x$$. Then the condition for $$\vec{a} \times \vec{b} = r$$ to follow is:

We have the two vectors

$$\vec a = 3\hat i + 2\hat j + x\hat k ,\qquad \vec b = \hat i - \hat j + \hat k ,$$

where $$x$$ is any real number. The question talks about $$\vec a \times \vec b = r$$. Because the left-hand side of this equation is a vector while $$r$$ is written as a scalar, the only way the statement can make sense is to interpret $$r$$ as the magnitude of the cross product:

$$r = \bigl\lvert\,\vec a \times \vec b\,\bigr\rvert.$$

To obtain the magnitude we must first find the cross product itself. Using the determinant form

$$ \vec a \times \vec b \;=\; \begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & 2 & x\\ 1 & -1 & 1 \end{vmatrix}, $$

we expand component by component:

$$ \vec a \times \vec b = \hat i\,(2\cdot1 \;-\; x\cdot(-1)) \;-\; \hat j\,(3\cdot1 \;-\; x\cdot1) \;+\; \hat k\,(3\cdot(-1) \;-\; 2\cdot1). $$

Now we simplify each bracket:

$$ \begin{aligned} \hat i &:& 2\cdot1 - x\cdot(-1) &= 2 + x,\\[4pt] -\hat j &:& 3\cdot1 - x\cdot1 &= 3 - x,\\[4pt] \hat k &:& 3\cdot(-1) - 2\cdot1 &= -3 - 2 = -5. \end{aligned} $$

So the cross-product vector itself is

$$ \vec a \times \vec b = (\,2 + x\,)\,\hat i \;-\;(3 - x)\,\hat j \;-\;5\,\hat k. $$

We now calculate its magnitude. For any vector $$P\hat i + Q\hat j + R\hat k$$ the magnitude formula is

$$ |\vec P| = \sqrt{P^{2} + Q^{2} + R^{2}}. $$

Here $$P = 2 + x,\; Q = -(3 - x),\; R = -5$$, so

$$ \bigl\lvert\,\vec a \times \vec b\,\bigr\rvert = \sqrt{(2 + x)^{2} + \bigl(-(3 - x)\bigr)^{2} + (-5)^{2}}. $$

We expand each square one by one:

$$ \begin{aligned} (2 + x)^{2} &= x^{2} + 4x + 4,\\[4pt] \bigl(-(3 - x)\bigr)^{2} &= (3 - x)^{2} = x^{2} - 6x + 9,\\[4pt] (-5)^{2} &= 25. \end{aligned} $$

Adding these three expressions gives

$$ x^{2} + 4x + 4 \;+\; x^{2} - 6x + 9 \;+\; 25 \;=\; 2x^{2} - 2x + 38. $$

Hence

$$ r = \bigl\lvert\,\vec a \times \vec b\,\bigr\rvert = \sqrt{\,2x^{2} - 2x + 38\,}. $$

To understand what numerical values $$r$$ can take, we must analyze the quadratic inside the square root:

$$ g(x) = 2x^{2} - 2x + 38. $$

This is an upward-opening parabola whose minimum value occurs at

$$ x = -\frac{b}{2a} = -\frac{-2}{2\cdot2} = \frac12. $$

Substituting $$x = \frac12$$ into $$g(x)$$ we obtain

$$ g\!\Bigl(\frac12\Bigr) = 2\!\left(\frac12\right)^{2} - 2\!\left(\frac12\right) + 38 = 2\!\left(\frac14\right) - 1 + 38 = \frac12 - 1 + 38 = \frac{75}{2}. $$

Therefore the smallest value that $$2x^{2} - 2x + 38$$ can take is $$\dfrac{75}{2}$$, and for all other values of $$x$$ the expression is larger. Since there is no upper bound on the quadratic (it grows to $$+\infty$$ as $$x \to \pm\infty$$), the set of possible values is

$$ [\,\tfrac{75}{2},\;\infty). $$

Taking square roots gives the complete range of $$r$$:

$$ r \in \Bigl[\,\sqrt{\tfrac{75}{2}},\;\infty\Bigr). $$

We simplify the radical once more:

$$ \sqrt{\tfrac{75}{2}} = \sqrt{\tfrac{25\cdot3}{2}} = 5\sqrt{\tfrac32}. $$

Thus the necessary and sufficient condition is

$$ r \;\ge\; 5\sqrt{\tfrac32}. $$

Among the printed alternatives, only Option B expresses exactly this inequality.

Hence, the correct answer is Option B.