If A(1,2,3), B(3,7,-2), C(6,7,7) and D(-1,0,- | vedclass

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(B) The centroid $G_1$ of $	riangle ABD$ is given by $\frac{A+B+D}{3} = \left(\frac{1+3-1}{3}, \frac{2+7+0}{3}, \frac{3-2-1}{3}ight) = (1, 3, 0)$.

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$\vec{r}=(1+t) \hat{i}+3 \hat{j}+3 t \hat{k}$

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(B) The centroid $G_1$ of $	riangle ABD$ is given by $\frac{A+B+D}{3} = \left(\frac{1+3-1}{3}, \frac{2+7+0}{3}, \frac{3-2-1}{3}ight) = (1, 3, 0)$. The position vector is $\vec{a} = \hat{i} + 3\hat{j}$.The centroid $G_2$ of $	riangle ACD$ is given by $\frac{A+C+D}{3} = \left(\frac{1+6-1}{3}, \frac{2+7+0}{3}, \frac{3+7-1}{3}ight) = (2, 3, 3)$. The position vector is $\vec{b} = 2\hat{i} + 3\hat{j} + 3\hat{k}$.The vector equation of a line passing through points with position vectors $\vec{a}$ and $\vec{b}$ is $\vec{r} = \vec{a} + t(\vec{b} - \vec{a})$.Calculating the direction vector: $\vec{b} - \vec{a} = (2\hat{i} + 3\hat{j} + 3\hat{k}) - (\hat{i} + 3\hat{j}) = \hat{i} + 3\hat{k}$.Substituting into the equation: $\vec{r} = (\hat{i} + 3\hat{j}) + t(\hat{i} + 3\hat{k}) = (1+t)\hat{i} + 3\hat{j} + 3t\hat{k}$.

If $A(1,2,3), B(3,7,-2), C(6,7,7)$ and $D(-1,0,-1)$ are points in a plane, then the vector equation of the line passing through the centroids of $\triangle ABD$ and $\triangle ACD$ is

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