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(B) Let the moving point be $P(x, y, z)$.Given that the distance from $P$ to $A(3, 4, -2)$ is equal to the distance from $P$ to $B(2, 3, -3)$.So,$PA^2

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$A$ plane whose normal is equally inclined to axes

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(B) Let the moving point be $P(x, y, z)$.Given that the distance from $P$ to $A(3, 4, -2)$ is equal to the distance from $P$ to $B(2, 3, -3)$.So,$PA^2 = PB^2$.$(x - 3)^2 + (y - 4)^2 + (z + 2)^2 = (x - 2)^2 + (y - 3)^2 + (z + 3)^2$Expanding both sides:$(x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 + 4z + 4) = (x^2 - 4x + 4) + (y^2 - 6y + 9) + (z^2 + 6z + 9)$Canceling $x^2, y^2, z^2$ from both sides:$-6x - 8y + 4z + 29 = -4x - 6y + 6z + 22$Rearranging the terms:$2x + 2y + 2z = 7$This is the equation of a plane. The normal vector to this plane is $\vec{n} = 2\hat{i} + 2\hat{j} + 2\hat{k}$.Since the components of the normal vector are equal,the normal is equally inclined to the coordinate axes.

$A$ point moves such that its distances from the points $(3, 4, -2)$ and $(2, 3, -3)$ remain equal. The locus of the point is

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