Answer is (A) → (r), (B) → (p), (C) → (q), (D) → (s)
Given abc/6 = 32, where A, B, C are, respectively, (a, 0, 0), (0, b, 0), (0, 0, c).
(A) Centroid of tetrahedron (x, y z) ≡ (a/4, b/4, c/4)
⇒ a = 4x, b = 4y, c = 4z
Therefore,
64xyz = 32 × 6 ⇒ xyz = 3
(B) Equidistant point (x, y, z) ≡ (a/2, b/2, c/2)
⇒ a = 2x, b = 2y, c = 2z
Therefore,
8abg = 32 × 6 ⇒ xyz = 24
(C) The equation of the plane is x/a + y/b + z/c = 1
Therefore, foot of perpendicular from the origin ≡ (x, y, z)
Therefore,
xyz = 6 × 32
⇒ (x2 + y2 + z2)3 = 192 × 8xyz = 1536xyz