Final answer:
The locus of the centroid of triangle ABC formed by a variable plane that remains at a constant distance 3p from the origin is 1/x + 1/y + 1/z = 1/3p.
Step-by-step explanation:
The student's question pertains to finding the locus of the centroid of a triangle formed by a plane intersecting the coordinate axes in three-dimensional space. The plane is at a constant distance 3p from the origin. This implies that the plane's equation, given by Ax + By + Cz = D, can be written under the condition A^2 + B^2 + C^2 = D^2, and given D is 3p, we have A^2 + B^2 + C^2 = (3p)^2. This is due to the fact that the distance from the origin to the plane (the perpendicular distance) is constant and represented by the intersection points with axes A, B, C, being located at (A,0,0), (0,B,0), and (0,0,C) respectively.
Since the centroid G of the triangle ABC is the average of the coordinates of A, B, and C, the coordinates of G are (A/3, B/3, C/3). To find the locus, we need an equation involving the coordinates of the centroid that holds for all positions of this centroid as the plane moves. The centroid equation thus becomes 1/(A/3) + 1/(B/3) + 1/(C/3) = 1/p which simplifies to 1/x + 1/y + 1/z = 1/3p.
Therefore, the correct answer is 1/x + 1/y + 1/z = 1/3p.