Question

a+b+c=5 and a²+b²+c²=29 then the value of 3ab+3bc+3ca is

Answer 1

$\\ \\ a+b+c = 5 \\ \\ {(a+b+c)}^{2} = {5}^{2} \\ \\ {a}^{2} + {b}^{2} + {c}^{2} + 2(ab+bc+ca) = 25 \\ \\ 2(ab+bc+ca) = -4 \\ \\ ab + bc + ca = -2 \\ \\ 3(ab+bc+ca) = 3 \times -2 \\ \\ 3ab + 3bc + 3ca = -6$