Match the following I. The centroid of the triangle formed by $(2, 3, -1), (5, 6, 3), (2, -3, 1)$ is a) $(2, 2, 2)$ II. The circumcentre of the triangle formed by $(1, 2, 3), (2, 3, 1), (3, 1, 2)$ is b) $(3, 1, 4)$ III. The orthocenter of the triangle formed by $(2, 1, 5), (3, 2, 3), (4, 0, 4)$ is c) $(1, 1, 0)$ IV. The incentre of the triangle formed by $(0, 0, 0), (3, 0, 0), (0, 4, 0)$ is d) $(3, 2, 1)$ e) $(0, 0, 0)$
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$I - d, II - a, III - b, IV - c$
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$I - a, II - b, III - c, IV - d$
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$I - d, II - e, III - b, IV - c$
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$I - d, II - a, III - e, IV - c$
A
Correct answer
Explanation
Centroid of (2,3,-1), (5,6,3), (2,-3,1) is ((2+5+2)/3, (3+6-3)/3, (-1+3+1)/3) = (3, 2, 1). This matches I-d. Checking other options confirms the mapping.
AI explanation
Using the coordinate geometry centroid formula ((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3), the centroid for the first set of points is ((2+5+2)/3, (6-3+3)/3, (3-1+1)/3), which equals (3, 2, 1). The points in the second set are equidistant from each other, making it an equilateral triangle where the circumcenter, orthocenter, and centroid coincide at (2, 2, 2). For the third set, using the orthocenter properties or verifying altitudes confirms the point is (3, 1, 4). For the fourth set, calculating side lengths gives 3, 4, and 5, making it a right-angled triangle at the origin, so the orthocenter is (0, 0, 0), and the standard incentre formula confirms the match is with option A.