Given:
\[
x y+y z+z x=0
\]

We have the expression to evaluate:

\[
\begin{aligned}
& \frac{1}{x^2-y z}+\frac{1}{y^2-x z}+\frac{1}{z^2-x y} \\
& \frac{1}{x^2-y z}=\frac{1}{x^2-(-x y-x z)}=\frac{1}{x(x+y+z)} \\
& \frac{1}{y^2-x z}=\frac{1}{y^2-(-x y-y z)}=\frac{1}{y(x+y+z)} \\
& \frac{1}{z^2-x y}=\frac{1}{z^2-(-y z-x z)}=\frac{1}{z(x+y+z)} \\
& \text { so } \frac{1}{x^2-y z}+\frac{1}{y^2-x z}+\frac{1}{z^2-x y} \\
& =\frac{1}{x(x+y+z)}+\frac{1}{y(x+y+z)}+\frac{1}{z(x+y+z)} \\
& =\frac{1}{x+y+z} \times\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \\
& =\frac{(x y+yz+zx)}{x y z(x+y+z)}=0
\end{aligned}
\]

Given the values for \(a\), \(b\), and \(c\) as:

– \(a = \sqrt{6} – \sqrt{5}\)
– \(b = \sqrt{5} – 2\)
– \(c = 2 – \sqrt{3}\)

And using the approximate square root values:

– \(\sqrt{6} \approx 2.45\)
– \(\sqrt{5} \approx 2.24\)
– \(\sqrt{3} \approx 1.73\)

We can approximate the values of \(a\), \(b\), and \(c\) as:

– \(a \approx 2.45 – 2.24 = 0.21\)
– \(b \approx 2.24 – 2 = 0.24\)
– \(c \approx 2 – 1.73 = 0.27\)

Thus, we have the order:

\[
a < b < c
\]

Therefore, the correct alternative among the given options is:

(a) \(a < b < c\)

Given \(x = \sqrt{3} + \sqrt{2}\), to find the value of \(x^3 – \frac{1}{x^3}\), we can proceed by calculating \(x^3\) and \(\frac{1}{x^3}\) separately.

First, calculate \(x^3\):

\[
x^3 = (\sqrt{3} + \sqrt{2})^3
\]

Applying the binomial expansion, we get:

\[
(\sqrt{3})^3 + 3(\sqrt{3})^2(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^2 + (\sqrt{2})^3
\]

\[
= 3\sqrt{3} + 3 \times 3 \sqrt{2} + 3 \times 2\sqrt{3} + 2\sqrt{2}
\]

\[
= 3\sqrt{3} + 9\sqrt{2} + 6\sqrt{3} + 2\sqrt{2}
\]

\[
= 9\sqrt{3} + 11\sqrt{2}
\]

Now, for \(\frac{1}{x^3}\), consider \(\frac{1}{x}\) first. Knowing that \(x = \sqrt{3} + \sqrt{2}\), we find the conjugate to rationalize the denominator for \(\frac{1}{x}\):

\[
\frac{1}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} – \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} – \sqrt{2})}
\]

\[
= \frac{\sqrt{3} – \sqrt{2}}{3 – 2}
\]

\[
= \sqrt{3} – \sqrt{2}
\]

Thus, \(\frac{1}{x} = \sqrt{3} – \sqrt{2}\), and hence \(\frac{1}{x^3} = (\sqrt{3} – \sqrt{2})^3\).

To simplify this, apply the binomial expansion similarly:

\[
(\sqrt{3} – \sqrt{2})^3 = (\sqrt{3})^3 – 3(\sqrt{3})^2(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^2 – (\sqrt{2})^3
\]

\[
= 3\sqrt{3} – 3 \times 3 \sqrt{2} + 3 \times 2\sqrt{3} – 2\sqrt{2}
\]

\[
= 9\sqrt{3} – 11\sqrt{2}
\]

Now, to find \(x^3 – \frac{1}{x^3}\):

\[
x^3 – \frac{1}{x^3} = (9\sqrt{3} + 11\sqrt{2}) – (9\sqrt{3} – 11\sqrt{2})
\]

\[
= 9\sqrt{3} + 11\sqrt{2} – 9\sqrt{3} + 11\sqrt{2}
\]

\[
= 22\sqrt{2}
\]

Therefore, the value of \(x^3 – \frac{1}{x^3}\) is \(22\sqrt{2}\), making the correct answer:

(c) \(22 \sqrt{2}\).

The temperature increased from \(21^{\circ}C\) to \(36^{\circ}C\) over a 5-hour period from 9:00 AM to 2:00 PM, which means the temperature rose \(36^{\circ}C – 21^{\circ}C = 15^{\circ}C\) in total.

To find the rate of increase per hour, divide the total temperature increase by the number of hours:

\[
\frac{15^{\circ}C}{5 \text{ hours}} = 3^{\circ}C/\text{hour}
\]

From 9:00 AM to noon (12:00 PM) is 3 hours. At a rate of \(3^{\circ}C/\text{hour}\), the temperature increase from 9:00 AM to noon would be:

\[
3 \text{ hours} \times 3^{\circ}C/\text{hour} = 9^{\circ}C
\]

Therefore, the temperature at noon, starting from \(21^{\circ}C\) at 9:00 AM, would be:

\[
21^{\circ}C + 9^{\circ}C = 30^{\circ}C
\]

So, the temperature at noon was \(30^{\circ}C\).

The correct answer is (b) \(30^{\circ} \mathrm{C}\).

To simplify the given expression \(\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\), let’s start by simplifying each square root by expressing them in terms of their prime factors and then simplify:

\[
\sqrt{32} = \sqrt{2^5} = \sqrt{16 \cdot 2} = 4\sqrt{2}
\]
\[
\sqrt{48} = \sqrt{2^4 \cdot 3} = 4\sqrt{3}
\]
\[
\sqrt{8} = \sqrt{2^3} = \sqrt{4 \cdot 2} = 2\sqrt{2}
\]
\[
\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}
\]

Substituting these simplified forms back into the original expression gives:

\[
\frac{4\sqrt{2} + 4\sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}
\]

We can take a 2 out of both the numerator and the denominator to simplify further:

\[
= \frac{2(2\sqrt{2} + 2\sqrt{3})}{2(\sqrt{2} + \sqrt{3})}
\]

Upon simplification, we find:

\[
= \frac{2\sqrt{2} + 2\sqrt{3}}{\sqrt{2} + \sqrt{3}}
\]

Since the terms in the numerator are double those in the denominator, simplifying the expression incorrectly suggested the terms could cancel out directly. The correct simplification should consider factoring correctly and assessing the common terms. The step where the 2 is factored out both in the numerator and denominator simplifies the expression directly, leading to:

\[
= 2
\]

Thus, the simplified value of the expression \(\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\) is indeed \(2\).

The correct answer is (c) 2.