$16a^2 + 4b^2 -9c^2 - 16ab - 12 bc - 24 ca = ?$

$(4a - 2b + 3c)^2$ $(4a - 2b - 3c)^2$ $(4a + 2b - 3c)^2$ $(4a + 2b + 3c)^2$

$(4a - 2b - 3c)^2$

We know that, (a + b + c)² = a² + b² + c² + 2(ab + bc + ca) = 16a2 + 4b2 + 9c2 - 16ab + 12bc - 24ca  = (4a)² + (-2b)² +(-3c)² +2{ 4a(-2b) + (-2b)(-3c) + 4a(-3c) } = (4a - 2b - 3c)²