JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

From |z - 6| = 5, point z lies on circle centered at (6, 0) with radius 5. From |z + 2 - 6i| = 5, point lies on circle centered at (-2, 6) with radius 5. Let z = x + iy. First condition gives: x^2 + y^2 = 12x - 11. Second gives: x^2 + y^2 = -4x + 12y - 15. Setting equal: 16x - 12y + 4 = 0, so x = 3y - 1 4 . Substituting into (x-6)^2 + y^2 = 25: (3y - 25)^2 + 16y^2 = 400, which simplifies to 25y^2 - 150y + 225 = 0, giving y = 3 and x = 2. Thus z = 2 + 3i, so z^2 = -5 + 12i and z^3 = -46 + 9i. z^3 + 3z^2 - 15z + 141 = (-46 + 9i) + 3(-5 + 12i) - 15(2 + 3i) + 141 = 50