A) (0, 1, 2)
B) (1, 3, 4)
C) (-1, 3, 4)
D) None of these
Correct Answer: B
Solution :
The equation of a line through the centre \[\hat{j}=2\hat{k}\] and normal to the given plane is \[r=\hat{j}+2\hat{k}+\lambda \,(\hat{i}+2\hat{j}+2\hat{k})\] ... (i) This meets the plane at a point for which we must have \[[(\hat{j}+2\hat{k})+\lambda (\hat{i}+2\hat{j}+2\hat{k})].\,\,(\hat{i}+2\hat{j}+2\hat{k})=15\] \[2(1+2\lambda )+\lambda +2(2+\lambda )=15\] \[\Rightarrow \] \[6+9\lambda =15\] \[\Rightarrow \] \[\lambda =1\] On putting \[\lambda =1\] in Eq. (i), we get \[r=\hat{i}+3\hat{j}+4\hat{k}\] \[\therefore \] Centre of the circle is (1, 3, 4).
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