If the coordinates of points A and B are (3, | vedclass

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(A) Let the coordinates of point $P$ be $(x, y, z)$.Given $A = (3, 4, 5)$ and $B = (-1, 3, -7)$.The square of the distance $PA$ is $PA^2 = (x - 3)^2 +

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$8x + 2y + 24z = 2k^2 - 9$

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(A) Let the coordinates of point $P$ be $(x, y, z)$.Given $A = (3, 4, 5)$ and $B = (-1, 3, -7)$.The square of the distance $PA$ is $PA^2 = (x - 3)^2 + (y - 4)^2 + (z - 5)^2 = x^2 - 6x + 9 + y^2 - 8y + 16 + z^2 - 10z + 25 = x^2 + y^2 + z^2 - 6x - 8y - 10z + 50$.The square of the distance $PB$ is $PB^2 = (x + 1)^2 + (y - 3)^2 + (z + 7)^2 = x^2 + 2x + 1 + y^2 - 6y + 9 + z^2 + 14z + 49 = x^2 + y^2 + z^2 + 2x - 6y + 14z + 59$.Substituting these into the equation $PA^2 - PB^2 + 2k^2 = 0$:$(x^2 + y^2 + z^2 - 6x - 8y - 10z + 50) - (x^2 + y^2 + z^2 + 2x - 6y + 14z + 59) + 2k^2 = 0$.Simplifying the expression:$(-6x - 2x) + (-8y + 6y) + (-10z - 14z) + (50 - 59) + 2k^2 = 0$.$-8x - 2y - 24z - 9 + 2k^2 = 0$.Rearranging the terms:$8x + 2y + 24z = 2k^2 - 9$.

If the coordinates of points $A$ and $B$ are $(3, 4, 5)$ and $(-1, 3, -7)$ respectively,find the equation of the locus of point $P(x, y, z)$ such that $PA^2 - PB^2 + 2k^2 = 0$.

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