A) (0, 1, 2)
B) (1, 3, 4)
C) (-1, 3, 4)
D) none of these
Correct Answer: B
Solution :
[b] The equation of the line through the centre \[\hat{j}+2\hat{k}\] and normal to the given plane is \[\vec{r}=\hat{j}+2\hat{k}+\lambda (\hat{i}+2\hat{j}+2\hat{k})\] ...(i) This meets the plane for which \[[\hat{j}+2\hat{k}+\lambda (\hat{i}+2\hat{j}+2\hat{k})]\cdot (\hat{i}+2\hat{j}+2\hat{k})=15\] Or \[6+9\lambda =15\] or \[\lambda =1\] Putting in (i) we get \[\vec{r}=\hat{j}+2\hat{k}+(\hat{i}+2\hat{j}+2\hat{k})\] \[=\hat{i}+3\hat{j}+4\hat{k}\] Hence, centre is\[(1,3,4)\].
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