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If $2 \hat{i}+\hat{j}-\hat{k}, \hat{i}-3 \hat{j}+5 \hat{k}$ and $-3 \hat{i}+4 \hat{j}+4 \hat{k}$ are the position vectors of three points A, B and C respectively, then
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The correct answer is:
$\mathrm{ABC}$ is a scalene triangle
$\begin{aligned}
\text { (d) } \overrightarrow{O A}=2 \hat{i}+\hat{j}-\hat{k}, \overrightarrow{O B}=\hat{i}-3 \hat{j}+5 \hat{k}, O C=-3 \hat{i}+4 \hat{j}+4 \hat{k} \\
\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=-\hat{i}+4 \hat{j}+6 \hat{k} \Rightarrow|\overrightarrow{A B}|=\sqrt{1+16+36}=\sqrt{53} \\
\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B}=-4 \hat{i}+7 \hat{j}-\hat{k} \Rightarrow|\overrightarrow{B C}|=\sqrt{16+49+1}=\sqrt{66} \\
\overrightarrow{C A}=\overrightarrow{O A}-\overrightarrow{O C}=5 \hat{i}-3 \hat{j}-5 \hat{k} \Rightarrow|\overrightarrow{C A}|=\sqrt{25+9+25}=\sqrt{59} \\
\quad|\overrightarrow{A B}| \neq|\overrightarrow{B C}| \neq|\overrightarrow{C A}|
\end{aligned}$
$\therefore \triangle A B C$ is a scalene triangle.
\text { (d) } \overrightarrow{O A}=2 \hat{i}+\hat{j}-\hat{k}, \overrightarrow{O B}=\hat{i}-3 \hat{j}+5 \hat{k}, O C=-3 \hat{i}+4 \hat{j}+4 \hat{k} \\
\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=-\hat{i}+4 \hat{j}+6 \hat{k} \Rightarrow|\overrightarrow{A B}|=\sqrt{1+16+36}=\sqrt{53} \\
\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B}=-4 \hat{i}+7 \hat{j}-\hat{k} \Rightarrow|\overrightarrow{B C}|=\sqrt{16+49+1}=\sqrt{66} \\
\overrightarrow{C A}=\overrightarrow{O A}-\overrightarrow{O C}=5 \hat{i}-3 \hat{j}-5 \hat{k} \Rightarrow|\overrightarrow{C A}|=\sqrt{25+9+25}=\sqrt{59} \\
\quad|\overrightarrow{A B}| \neq|\overrightarrow{B C}| \neq|\overrightarrow{C A}|
\end{aligned}$
$\therefore \triangle A B C$ is a scalene triangle.
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