Let →a,→b,→c are vectors such that |→a|=1,|→b|=6,|→c|=2√3 and (→a+2→b) is perpendicular to →c,(→b+2→c) is perpendicular to →a and (→c+2→a) is perpendicular to →b. Then the value of |→a+→b+→c|=

Let →a,→b,→c are vectors such that |→a|=1,|→b|=6,|→c|=2√

1.	Let →a,→b,→c are vectors such that \|→a\|=1,\|→b\|=6,\|→c\|=2√3 and (→a+2→b) is perpendicular to →c,(→b+2→c) is perpendicular to →a and (→c+2→a) is perpendicular to →b. Then the value of \|→a+→b+→c\|=
Answer» Let $\to a, \to b, \to c$ are vectors such that $\| \to a \| = 1, \| \to b \| = 6, \| \to c \| = 2 \sqrt{3}$ and $(\to a + 2 \to b)$ is perpendicular to $\to c, (\to b + 2 \to c)$ is perpendicular to $\to a$ and $(\to c + 2 \to a)$ is perpendicular to $\to b .$ Then the value of $\| \to a + \to b + \to c \| =$