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

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(B) Let the position vectors of vertices $A, B, C$ be $\vec{a} = 3\hat{i}-2\hat{j}-\hat{k}$,$\vec{b} = -2\hat{i}-\hat{j}+3\hat{k}$,and $\vec{c} = -\ha

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$\overrightarrow{0}$

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(B) Let the position vectors of vertices $A, B, C$ be $\vec{a} = 3\hat{i}-2\hat{j}-\hat{k}$,$\vec{b} = -2\hat{i}-\hat{j}+3\hat{k}$,and $\vec{c} = -\hat{i}+3\hat{j}-2\hat{k}$.First,calculate the side lengths squared:$AB^2 = |\vec{b}-\vec{a}|^2 = |(-5)\hat{i} + \hat{j} + 4\hat{k}|^2 = 25+1+16 = 42$$BC^2 = |\vec{c}-\vec{b}|^2 = |\hat{i} + 4\hat{j} - 5\hat{k}|^2 = 1+16+25 = 42$$AC^2 = |\vec{c}-\vec{a}|^2 = |(-4)\hat{i} + 5\hat{j} - \hat{k}|^2 = 16+25+1 = 42$Since $AB^2 = BC^2 = AC^2 = 42$,the triangle is equilateral.For an equilateral triangle,the orthocenter $H$ coincides with the centroid $G$.The centroid $G$ is given by $\frac{\vec{a}+\vec{b}+\vec{c}}{3} = \frac{(3-2-1)\hat{i} + (-2-1+3)\hat{j} + (-1+3-2)\hat{k}}{3} = \vec{0}$.Thus,$H = (0,0,0)$.Then,$\overrightarrow{HA} + \overrightarrow{HB} + \overrightarrow{HC} = (\vec{a}-\vec{h}) + (\vec{b}-\vec{h}) + (\vec{c}-\vec{h}) = \vec{a}+\vec{b}+\vec{c} - 3\vec{h}$.Since $\vec{h} = \vec{0}$ and $\vec{a}+\vec{b}+\vec{c} = \vec{0}$,we have $\overrightarrow{HA} + \overrightarrow{HB} + \overrightarrow{HC} = \vec{0}$.

$3 \hat{i}-2 \hat{j}-\hat{k}, -2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle ABC$ respectively. If $H$ is its orthocenter,then $\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC} = $

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