Let vec{i}+vec{j}+vec{k}, a_1 vec{i}+b_1 vec{ | vedclass

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(A) Let the position vectors of points $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Given $\vec{a} = \vec{i}+\vec{j}+\vec{k}$.The

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$3$

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(A) Let the position vectors of points $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Given $\vec{a} = \vec{i}+\vec{j}+\vec{k}$.The centroid of the triangular face $BCD$ is given by $\vec{G}_{BCD} = \frac{\vec{b}+\vec{c}+\vec{d}}{3} = \frac{2}{3}(\vec{i}+\vec{j}+\vec{k})$.Thus, $\vec{b}+\vec{c}+\vec{d} = 2(\vec{i}+\vec{j}+\vec{k})$.The centroid of the tetrahedron $ABCD$ is $\vec{G} = \frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}$.Substituting the values, $\vec{G} = \frac{(\vec{i}+\vec{j}+\vec{k}) + 2(\vec{i}+\vec{j}+\vec{k})}{4} = \frac{3}{4}(\vec{i}+\vec{j}+\vec{k}) = \frac{3}{4}\vec{i}+\frac{3}{4}\vec{j}+\frac{3}{4}\vec{k}$.Comparing with $\alpha \vec{i}+\beta \vec{j}+\gamma \vec{k}$, we get $\alpha = \frac{3}{4}, \beta = \frac{3}{4}, \gamma = \frac{3}{4}$.Then $2\alpha+\beta+\gamma = 2(\frac{3}{4}) + \frac{3}{4} + \frac{3}{4} = \frac{6}{4} + \frac{6}{4} = \frac{12}{4} = 3$.

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