Let’s denote the vectors as \(\vec{a}, \vec{b}, \) and \(\vec{c}\).

Volume of the First Parallelepiped

The volume of the parallelepiped formed by vectors \(\vec{a} \times \vec{b}, \vec{b} \times \vec{c},\) and \(\vec{c} \times \vec{a}\) can be expressed as the scalar triple product of these vectors:

\[
\text{Volume} = [(\vec{a} \times \vec{b}) \cdot ((\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}))]
\]

Using the vector triple product identity, \(\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) – \vec{C}(\vec{A} \cdot \vec{B})\), we can simplify the expression:

\[
\text{Volume} = [(\vec{a} \times \vec{b}) \cdot (\vec{b}(\vec{b} \cdot \vec{a}) – \vec{a}(\vec{b} \cdot \vec{b}))]
\]

Expanding further:

\[
\text{Volume} = [(\vec{a} \times \vec{b}) \cdot \vec{b}(\vec{b} \cdot \vec{a})] – [(\vec{a} \times \vec{b}) \cdot \vec{a}(\vec{b} \cdot \vec{b})]
\]

Since \((\vec{a} \times \vec{b}) \cdot \vec{b} = 0\) and \((\vec{a} \times \vec{b}) \cdot \vec{a} = 0\) (as the cross product is perpendicular to both vectors), the volume of the first parallelepiped is 0.

Volume of the Second Parallelepiped

For the second parallelepiped, we have the vectors \((\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c})\), \((\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})\), and \((\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b})\). The volume can be similarly expressed as the scalar triple product:

\[
\text{Volume} = [((\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c})) \cdot (((\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})) \times ((\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b})))]
\]

Using the same vector triple product identity and the fact that the cross product of any two vectors is perpendicular to both, we can deduce that the volume of the second parallelepiped is also 0.

In summary, the volume of the second parallelepiped with \((\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c})\), \((\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})\), and \((\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b})\) on coterminal edges is \(0 \, \mathrm{cu. \, units}\).

Given the system of equations:

1. \(a + b + c = 3\)
2. \(a^2 + b^2 + c^2 = 6\)
3. \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\)

We want to find the value of \(abc\).

To solve this, let’s first analyze the given information. The third equation can be written as:

\[
\frac{ab + ac + bc}{abc} = 1
\]

From this, we get:

\[
ab + ac + bc = abc
\]

From the first equation, we know the sum of \(a\), \(b\), and \(c\). To find \(ab + ac + bc\), we can square the first equation and compare it to the second equation. Squaring the first equation gives us:

\[
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) = 3^2 = 9
\]

Given \(a^2 + b^2 + c^2 = 6\) from the second equation, we can substitute that into our squared first equation to find \(ab + ac + bc\):

\[
6 + 2(ab + ac + bc) = 9
\]

Solving for \(ab + ac + bc\):

\[
2(ab + ac + bc) = 3
\]
\[
ab + ac + bc = \frac{3}{2}
\]

Given \(ab + ac + bc = abc\), we have:

\[
abc = \frac{3}{2}
\]

Therefore, the value of \(abc\) is \(\frac{3}{2}\).

Answer: (b) \(\frac{3}{2}\).

Given

\[
\begin{aligned}
& x=3+2 \sqrt{2} \\
& x=2+1+2 \sqrt{2} \\
& x=(\sqrt{2})^2+(1)^2+2.1 \cdot \sqrt{2} \\
& x=(\sqrt{2}+1)^2 \\
& \sqrt{x}=(\sqrt{2}+1) \\
& \frac{1}{\sqrt{x}}=\frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1
\end{aligned}
\]

Now, \(\sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{2}+1-(\sqrt{2}-1)=\sqrt{2}+1-\sqrt{2}+1\)
\[
\sqrt{x}-\frac{1}{\sqrt{x}}=2
\]

Therefore, Correct option is 2.

Given the equation \(a + b + c = 0\), we can solve for one of the variables in terms of the others. Without loss of generality, let’s solve for \(c\):

\[
c = -a – b
\]

Substituting this into the given expression:

\[
\frac{a+b}{c} – \frac{2b}{c+a} + \frac{b+c}{a} = \frac{a+b}{-a-b} – \frac{2b}{-a-b+a} + \frac{b-a-b}{a}
\]

This simplifies further to:

\[
\frac{a+b}{-a-b} – \frac{2b}{-b} + \frac{-a}{a}
\]

Since \(\frac{a+b}{-a-b} = -1\) (because the numerator is the negation of the denominator), and \(\frac{-a}{a} = -1\), the expression further simplifies to:

\[
-1 + \frac{2b}{b} – 1 = -1 + 2 – 1
\]

\[
= 2 – 2 = 0
\]

Thus, the value of the given expression, under the condition that \(a + b + c = 0\), is \(0\).

Answer: (a) 0.