A) \[\vec{r}.(20\,\hat{i}-8\hat{j}+5\hat{k})=120\]
B) \[\vec{r}.(20\,\hat{i}-8\hat{j}+5\hat{k})=1\]
C) \[\vec{r}.(20\,\hat{i}-8\hat{j}+5\hat{k})=2\]
D) \[\vec{r}.(20\,\hat{i}-8\hat{j}+5\hat{k})=20\]
Correct Answer: A
Solution :
Let the coordinates of P, Q and R are \[(a,0,0),\,\,(0,b,0)\] and \[(0,0,c)\] respectively. Given, centroid \[=(2,-5,8)\] \[\therefore \] \[\frac{(a+0+0)}{3}=2\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,a=6\] \[\frac{0+b+0}{3}=-5\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,b=-15\] and \[\frac{0+0+c}{3}=8\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,c=24\] \[\therefore \] Position vector of P,Q and R are \[\vec{a}=(6\hat{i}+0\hat{j}+0\hat{k}),\] \[\vec{b}=(0\hat{i}-15\hat{j}+0\hat{k})\] and \[\vec{c}=(0\hat{i}+0\hat{j}+24\hat{k})\] \[\therefore \] Equation of plane is \[\vec{r}.(\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a})=[\vec{a}\vec{b}\vec{c}]\] \[\Rightarrow \] \[\vec{r}.(-90\hat{k}-360\hat{i}+144\hat{j})=-2160\] \[\Rightarrow \] \[\vec{r}.(5\hat{k}+20\hat{i}-8\hat{j})=120\] \[\Rightarrow \] \[\vec{r}.(20\hat{i}-8\hat{j}+5\hat{k})=120\]
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