If the volume of a parallelepiped, whose coterminous edges are given by the vectors $$\vec{a} = \hat{i} + \hat{j} + n\hat{k}$$, $$\vec{b} = 2\hat{i} + 4\hat{j} - n\hat{k}$$ and $$\vec{c} = \hat{i} + n\hat{j} + 3\hat{k}$$ $$(n \geq 0)$$ is 158 cubic units, then:

For a parallelepiped whose coterminous edges are represented by the vectors $$\vec a,\;\vec b,\;\vec c,$$ the volume is given by the absolute value of the scalar triple product, that is

$$V \;=\; \bigl|\,\vec a \cdot (\vec b \times \vec c)\bigr|.$$

We have

$$\vec a = \hat i + \hat j + n\hat k \;=\; (1,\,1,\,n),$$ $$\vec b = 2\hat i + 4\hat j - n\hat k \;=\; (2,\,4,\,-n),$$ $$\vec c = \hat i + n\hat j + 3\hat k \;=\; (1,\,n,\,3).$$

First we find the cross product $$\vec b \times \vec c.$$ Using the determinant form

$$\vec b \times \vec c \;=\; \begin{vmatrix} \hat i & \hat j & \hat k\\[4pt] 2 & 4 & -n\\[4pt] 1 & n & 3 \end{vmatrix},$$

we expand:

$$$ \vec b \times \vec c = \hat i\,(4\cdot 3 - (-n)\cdot n) - \hat j\,(2\cdot 3 - (-n)\cdot 1) + \hat k\,(2\cdot n - 4\cdot 1). $$$

Evaluating each term we obtain

$$$ \vec b \times \vec c = \hat i\,(12 + n^{2}) - \hat j\,(6 + n) + \hat k\,(2n - 4) = \bigl(12 + n^{2},\, -\!(6 + n),\, 2n - 4\bigr). $$$

Now we take the dot product with $$\vec a = (1,\,1,\,n):$$

$$$ \vec a \cdot (\vec b \times \vec c) = 1\,(12 + n^{2}) + 1\,\bigl(-6 - n\bigr) + n\,(2n - 4). $$$

Carrying out the arithmetic step by step:

$$$ 1\,(12 + n^{2}) \;=\; 12 + n^{2}, \quad 1\,\bigl(-6 - n\bigr) \;=\; -6 - n, \quad n\,(2n - 4) \;=\; 2n^{2} - 4n. $$$

Adding these three results gives

$$$ 12 + n^{2} \;+\; (-6 - n) \;+\; (2n^{2} - 4n) = 3n^{2} - 5n + 6. $$$

Hence

$$\vec a \cdot (\vec b \times \vec c) = 3n^{2} - 5n + 6.$$

By the volume formula we must have

$$\bigl|3n^{2} - 5n + 6\bigr| = 158.$$

This gives two algebraic possibilities:

1. $$3n^{2} - 5n + 6 = 158,$$  2. $$3n^{2} - 5n + 6 = -158.$$

For the first equation:

$$$ 3n^{2} - 5n + 6 = 158 \;\Longrightarrow\; 3n^{2} - 5n - 152 = 0. $$$

Using the quadratic formula $$n = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$ with $$a=3,\; b=-5,\; c=-152,$$ we have

$$$ n = \frac{5 \pm \sqrt{(-5)^{2} - 4\cdot 3 \cdot (-152)}}{2\cdot 3} = \frac{5 \pm \sqrt{25 + 1824}}{6} = \frac{5 \pm \sqrt{1849}}{6}. $$$

Since $$\sqrt{1849} = 43,$$ this becomes

$$$ n = \frac{5 \pm 43}{6}. $$$

Thus

$$$ n = \frac{5 + 43}{6} = \frac{48}{6} = 8 \quad\text{or}\quad n = \frac{5 - 43}{6} = \frac{-38}{6} = -\frac{19}{3}. $$$

Because the problem states $$n \ge 0,$$ we retain only

$$n = 8.$$

For the second equation $$3n^{2} - 5n + 6 = -158$$ we would have

$$$ 3n^{2} - 5n + 164 = 0, $$$

whose discriminant is $$(-5)^{2} - 4\cdot 3\cdot 164 = 25 - 1968 = -1943,$$ a negative number, so this equation has no real solutions. Hence the only admissible value is

$$n = 8.$$

Now we examine the statements offered in the options. We shall substitute $$n = 8$$ into each relevant expression.

Option A: $$\vec a \cdot \vec c = (1,\,1,\,8)\cdot(1,\,8,\,3) = 1\cdot 1 + 1\cdot 8 + 8\cdot 3 = 1 + 8 + 24 = 33.$$ The option claims $$17$$, so it is false.

Option B: $$\vec b \cdot \vec c = (2,\,4,\,-8)\cdot(1,\,8,\,3) = 2\cdot 1 + 4\cdot 8 + (-8)\cdot 3 = 2 + 32 - 24 = 10.$$ This exactly matches the value given in the option, so Option B is true.

Option C: states $$n = 7$$, which contradicts $$n = 8$$; therefore it is false.

Option D: states $$n = 9$$; again this contradicts $$n = 8$$, so it is false.

Only Option B withstands all checks.

Hence, the correct answer is Option B.