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**Hint:**To check the three points A, B and C given above are the vertices of a right angled triangle, first of all, we are going to find the distance between AB, BC and AC using the following formula: $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$. And then we will check whether the three sides AB, BC and AC are following Pythagoras theorem or not. In Pythagoras theorem, the square of the longest side is equal to the sum of squares of the other two sides.

**Complete step by step solution:**

We are going to find the distance of AB, BC and AC using the following formula:

$\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$

In the above formula, $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ is one of the vertices and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is the other vertex.

Now, finding the distance between A (3, 6, 9) and B (10, 20, 30) by considering point A as $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and B as $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ in the above formula we get,

$\begin{align}

& \Rightarrow \sqrt{{{\left( 10-3 \right)}^{2}}+{{\left( 20-6 \right)}^{2}}+{{\left( 30-9 \right)}^{2}}} \\

& =\sqrt{{{\left( 7 \right)}^{2}}+{{\left( 14 \right)}^{2}}+{{\left( 21 \right)}^{2}}} \\

& =\sqrt{49+196+441} \\

& =\sqrt{686} \\

& =26.2 \\

\end{align}$

The distance between B (10, 20, 30) and C (25, -41, 5) is calculated by considering point B as $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and C as $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$.

$\begin{align}

& \Rightarrow \sqrt{{{\left( 25-10 \right)}^{2}}+{{\left( -41-20 \right)}^{2}}+{{\left( 5-30 \right)}^{2}}} \\

& =\sqrt{{{\left( 15 \right)}^{2}}+{{\left( -61 \right)}^{2}}+{{\left( -25 \right)}^{2}}} \\

& =\sqrt{225+3721+625} \\

& =\sqrt{4571} \\

& =67.61 \\

\end{align}$

The distance between A (3, 6, 9) and C (25, -41, 5) is calculated by considering A as $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and B as $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$.

$\begin{align}

& \Rightarrow \sqrt{{{\left( 25-3 \right)}^{2}}+{{\left( -41-6 \right)}^{2}}+{{\left( 5-9 \right)}^{2}}} \\

& =\sqrt{{{\left( 22 \right)}^{2}}+{{\left( -47 \right)}^{2}}+{{\left( -4 \right)}^{2}}} \\

& =\sqrt{484+2209+16} \\

& =\sqrt{2709} \\

& =52.05 \\

\end{align}$

From the above we found the distances AB, BC and AC as follows:

$\begin{align}

& AB=26.2 \\

& BC=67.61 \\

& AC=52.05 \\

\end{align}$

As the largest distance is of BC so verifying Pythagoras theorem in the above we get,

$\begin{align}

& \Rightarrow {{\left( BC \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( AC \right)}^{2}} \\

& \Rightarrow {{\left( 67.61 \right)}^{2}}={{\left( 26.2 \right)}^{2}}+{{\left( 52.05 \right)}^{2}} \\

& \Rightarrow 4571.1=686.44+2709.2 \\

& \Rightarrow 4571.1=3395.64 \\

\end{align}$

**As you can see that L.H.S is not equal to R.H.S so the vertices A, B and C are not forming a right angled triangle.**

**Note:**The other thing is that while verifying the Pythagoras theorem, we have rejected that L.H.S is not equal to R.H.S because the difference between L.H.S and R.H.S is very large which you can see from the below:

$\begin{align}

& \Rightarrow {{\left( BC \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( AC \right)}^{2}} \\

& \Rightarrow {{\left( 67.61 \right)}^{2}}={{\left( 26.2 \right)}^{2}}+{{\left( 52.05 \right)}^{2}} \\

& \Rightarrow 4571.1=686.44+2709.2 \\

& \Rightarrow 4571.1=3395.64 \\

\end{align}$

As you can see, the difference between L.H.S and R.H.S is 1175.46 which is quite large that’s why we said that L.H.S is not equal to R.H.S. If the difference between L.H.S and R.H.S is very small like 0.11 or lesser than that then we can say that L.H.S is equal to R.H.S.

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