If hat{i}-2 hat{j}+3 hat{k},2 hat{i}+3 hat{j} | vedclass

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(D) Let the position vectors be $\vec{a} = \hat{i}-2 \hat{j}+3 \hat{k}$,$\vec{b} = 2 \hat{i}+3 \hat{j}+\hat{k}$,and $\vec{c} = -3 \hat{i}-\hat{j}-2 \h

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form a scalene triangle

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(D) Let the position vectors be $\vec{a} = \hat{i}-2 \hat{j}+3 \hat{k}$,$\vec{b} = 2 \hat{i}+3 \hat{j}+\hat{k}$,and $\vec{c} = -3 \hat{i}-\hat{j}-2 \hat{k}$.Calculating the vectors for the sides:$\overrightarrow{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (3-(-2))\hat{j} + (1-3)\hat{k} = \hat{i} + 5\hat{j} - 2\hat{k}$.$\overrightarrow{BC} = \vec{c} - \vec{b} = (-3-2)\hat{i} + (-1-3)\hat{j} + (-2-1)\hat{k} = -5\hat{i} - 4\hat{j} - 3\hat{k}$.$\overrightarrow{AC} = \vec{c} - \vec{a} = (-3-1)\hat{i} + (-1-(-2))\hat{j} + (-2-3)\hat{k} = -4\hat{i} + \hat{j} - 5\hat{k}$.Calculating the magnitudes:$|\overrightarrow{AB}| = \sqrt{1^2 + 5^2 + (-2)^2} = \sqrt{1 + 25 + 4} = \sqrt{30}$.$|\overrightarrow{BC}| = \sqrt{(-5)^2 + (-4)^2 + (-3)^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}$.$|\overrightarrow{AC}| = \sqrt{(-4)^2 + 1^2 + (-5)^2} = \sqrt{16 + 1 + 25} = \sqrt{42}$.Since all side lengths are different,the points form a scalene triangle.

If $\hat{i}-2 \hat{j}+3 \hat{k}$,$2 \hat{i}+3 \hat{j}+\hat{k}$,and $-3 \hat{i}-\hat{j}-2 \hat{k}$ are the position vectors of three points $A$,$B$,and $C$ respectively,then $A$,$B$,and $C$

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