A) \[\vec{r}\cdot (\hat{i}+2\hat{j}-3\hat{k})=0\]
B) \[\vec{r}\cdot (\hat{i}+4\hat{j}+\hat{k})=0\]
C) \[\vec{r}\cdot (\hat{i}+5\hat{j}-5\hat{k})=0\]
D) \[\vec{r}\cdot (\hat{i}+\hat{j}+3\hat{k})=0\]
Correct Answer: A
Solution :
The cartesian form of an equation of plane is \[x+3y-z=0\] and \[y+2z=0\] The line of intersection of two planes is \[(x+3y-z)+\lambda (y+2z)=0\] Since, it is passing through\[\left( -1,-1,-1 \right)\] \[\therefore \] \[(-1,\,-3+1)+\lambda (-1-2)=0\] \[\Rightarrow \] \[-\,3-3\lambda =0\] \[\Rightarrow \] \[\lambda =-1\] \[\therefore \]\[(x+3y-z)-1\,(y+2z=0\] \[\Rightarrow \] \[x+2y-3z=0\] Hence, equation of plane is \[\overrightarrow{r}\cdot (\hat{i}+2\hat{j}-3\hat{k})=0\]
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