The locus of the point $P$ such that $P A^{2} + P B^{2} = 10$ where $A = (2, 3, 4), B = (3, - 4, 2)$ is

x^{2} + y^{2} + z^{2} - x + y - 4 z + 12 = 0

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x^{2} + y^{2} + z^{2} - 5 x + y - 6 z + 24 = 0

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2 (x^{2} + y^{2} + z^{2}) - x + y - 4 z + 12 = 0

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x^{2} + y^{2} + z^{2} + x - y + 4 z - 12 = 0

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Solution

The correct option is B $x^{2} + y^{2} + z^{2} - 5 x + y - 6 z + 24 = 0$
Let $P = (x, y, z)$
$P A^{2} = (x - 2)^{2} + (y - 3)^{2} + (z - 4)^{2}$
$P B^{2} = (x - 3)^{2} + (y + 4)^{2} + (z - 2)^{2}$
$P A^{2} + P B^{2} = (x - 2)^{2} + (x - 3)^{2} + (y - 3)^{2} + (y + 4)^{2} + (z - 4)^{2} + (z - 2)^{2}$
$= x^{2} - 4 x + 4 + x^{2} - 6 x + 9 + y^{2} - 6 y + 9 + y^{2} + 8 y + 16 + z^{2} - 8 z + 16 + z^{2} - 4 z + 4$
$= 2 x^{2} - 10 x + 13 + 2 y^{2} + 2 y + 2 y + 25 + 2 z^{2} - 12 z + 20$
$2 x^{2} + 2 y^{2} + 2 z^{2} - 10 x + 2 y - 12 z + 58$
but $P A^{2} + P B^{2} = 10$
$∴ 2 x^{2} + 2 y^{2} + 2 z^{2} - 10 x + 2 y - 12 z + 48 = 0$
Dividing throughout by $2$ ,
$x^{2} + y^{2} + z^{2} - 5 x + y - 6 z + 24 = 0$ .

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