If 12 hat{i}-12 hat{j}-18 hat{k},-3 hat{i}-6 | vedclass

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(D) Let the position vectors of vertices $A, B, C$ be $\vec{a} = 12 \hat{i}-12 \hat{j}-18 \hat{k}$,$\vec{b} = -3 \hat{i}-6 \hat{j}-9 \hat{k}$,and $\ve

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$4 \hat{i}-5 \hat{j}-17 \hat{k}$

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(D) Let the position vectors of vertices $A, B, C$ be $\vec{a} = 12 \hat{i}-12 \hat{j}-18 \hat{k}$,$\vec{b} = -3 \hat{i}-6 \hat{j}-9 \hat{k}$,and $\vec{c} = 3 \hat{i}+3 \hat{j}-24 \hat{k}$.Calculate the side lengths:$a = |\vec{BC}| = |(3 - (-3))\hat{i} + (3 - (-6))\hat{j} + (-24 - (-9))\hat{k}| = |6\hat{i} + 9\hat{j} - 15\hat{k}| = \sqrt{6^2 + 9^2 + (-15)^2} = \sqrt{36 + 81 + 225} = \sqrt{342}$.$b = |\vec{AC}| = |(3 - 12)\hat{i} + (3 - (-12))\hat{j} + (-24 - (-18))\hat{k}| = |-9\hat{i} + 15\hat{j} - 6\hat{k}| = \sqrt{(-9)^2 + 15^2 + (-6)^2} = \sqrt{81 + 225 + 36} = \sqrt{342}$.$c = |\vec{AB}| = |(-3 - 12)\hat{i} + (-6 - (-12))\hat{j} + (-9 - (-18))\hat{k}| = |-15\hat{i} + 6\hat{j} + 9\hat{k}| = \sqrt{(-15)^2 + 6^2 + 9^2} = \sqrt{225 + 36 + 81} = \sqrt{342}$.Since $a=b=c$,the triangle is equilateral.The incentre of an equilateral triangle is the centroid,given by $\frac{\vec{a} + \vec{b} + \vec{c}}{3}$.Incentre $= \frac{(12-3+3)\hat{i} + (-12-6+3)\hat{j} + (-18-9-24)\hat{k}}{3} = \frac{12\hat{i} - 15\hat{j} - 51\hat{k}}{3} = 4\hat{i} - 5\hat{j} - 17\hat{k}$.

If $12 \hat{i}-12 \hat{j}-18 \hat{k}$,$-3 \hat{i}-6 \hat{j}-9 \hat{k}$ and $3 \hat{i}+3 \hat{j}-24 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ respectively of $\triangle ABC$,then the position vector of the incentre of $\triangle ABC$ is

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