The centre and radius of the sphere (r-3i-4j + 5k)·(r-2i + 3j-4k) = 0 are

Centre=3i+4j-5k+2i-3j+4k2 =5i+j-k2 2r=1+49+81=131

Centre=3i+4j-5k+2i-3j+4k2 =5i+j-k2 2r=1+49+81=131

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The centre and radius of the sphere $(r - 3 i - 4 j + 5 k) \cdot (r - 2 i + 3 j - 4 k) = 0$ are

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$5 i + j - k, \sqrt{139}$

$\frac{1}{2} (5 i + j - k), \frac{\sqrt{139}}{2}$

$i + 7 j - 9 k, \sqrt{27}$

$\frac{1}{2} (i + 7 j - 9 k), \sqrt{79}$

answer is B.

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$Centre = \frac{3 i + 4 j - 5 k + 2 i - 3 j + 4 k}{2} = \frac{5 i + j - k}{2} 2 r = \sqrt{1 + 49 + 81} = \sqrt{131}$

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