Q.

The point on z-axis  which is equidistant from the points (1,5,7) and (5,1,-4)

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a

0,0,32

b

0,0,1

c

0,0,132

d

0,0,5

answer is A.

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Detailed Solution

Suppose that P(0,0,z) is any point on z- axis  such that it is equidistant from the points  A(1,5,7)and B(5,1,-4) It gives PA=PB(z−7)2=(z+4)2​−14z+49=8z+16​     22z=33​     z=32Therefore, the point is P0,0,32
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