For A, let the roots be a-d, a, a+d; their sum is 3a = 9, so a = 3. The product is a(a^2-d^2) = 3(9-d^2) = 26, which yields k = 24 (matching 5). For B, let the roots be a/r, a, ar; their product is a^3 = 64, giving a = 4. The sum of roots taken two at a time is a^2/r + ar + a^2 = 16/r + 4r + 16 = k; since a/r + a + ar = 14, we get 4/r + 4r = 10, so 16/r + 4r = 40, meaning k = 56 (matching 1). For C, if roots are in H.P., the roots of the reciprocal equation -x^3 + 6x^2 - kx + 6 = 0 are in A.P.; letting them be a-d, a, a+d gives 3a = 6, so a = 2, and the product a(a^2-d^2) = 2(4-d^2) = 6 yields d^2 = 1. The sum of products two at a time is a^2-d^2 + a^2 + a^2-d^2 = 3a^2 - 2d^2 = 10, thus k = 11 and 3k = 33 (matching 4). For D, if two roots sum to 3, their product is -6 (from the constant term 6 divided by -1); substituting x = 3 into the equation gives 27 - 3k + 6 = 0, yielding k = 11; the product of all roots is -6, so the third root is -6 / 3 = -3 (matching 3). The correct matching is A-5, B-1, C-4, D-3.