Let (3, 4, -1) and (-1, 2, 3) are the end points of a diameter of sphere. Then the radius of the sphere is equal to
[Orissa JEE 2003]

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Solution

The correct option is C
3

$d = \sqrt{(- 1 - 3)^{2} + (2 - 4)^{2} + (3 + 1)^{2}} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$

So, radius $r = \frac{d}{2} = \frac{6}{2} = 3$

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