If a, b, c are the sides of the triangle ABC and $a^2, b^2, c^2$ are the roots of the equation $x^3-px^2+qx - \lambda =0$, then match the entries of col. I with those of col. II. Column-I Column-II (a) $\dfrac{cos A}{a} + \dfrac{cos B}{b} + \dfrac{cos C}{c} = \dfrac{1}{2}$ (p) $p=2, q=1, \lambda=4$ (b) $a cos A + b cos B + c cos C =0$ (q) $p=1, q=1/4, \lambda=1$ (c) $sin 2A + sin 2B + sin 2C =0$ (r) $p=3, q=9/4, \lambda=9$ (d) $a sin A + b sin B + c sin C = 2 \Delta$ (s) $p=4, q=4, \lambda=2$

By Vieta's formulas for the roots a^2, b^2, and c^2, we establish p = a^2 + b^2 + c^2, q = a^2*b^2 + b^2*c^2 + c^2*a^2, and lambda = a^2*b^2*c^2. For an equilateral triangle where a = b = c, these values become p = 3a^2, q = 3a^4, and lambda = a^6, which aligns with the ratios found in option (r) of p = 3, q = 9/4, and lambda = 9. If the triangle is right-angled at C, then a^2 + b^2 = c^2, making p = 2c^2 and lambda = a^2*b^2*c^2, which corresponds to the properties of a right triangle matching option (p). By applying trigonometric identities for triangles, condition (a) evaluates to 1 + q/p, matching the valid cases (p, q, r), while condition (d) simplifies correctly to the triangle's area, also matching (p, q, r). Conditions (b) and (c) are directly governed by the right-angle trigonometric identity and match all applicable geometric setups (p, q, r, s). Therefore, the entries match as (a), (d) mapping to (p, q, r) and (b), (c) mapping to (p, q, r, s).

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