`|vec(a)|=2,|vec(b)|=3,|vec(c)|=6` . Angle between `vec(a)` and `vec(b),vec(b)` and `vec(c)` and `vec(c)` and `vec(a)` is `120^(@)` each, find `|vec(a)+vec(b)+vec(c)|`?

Let vec a,vec b,vec c are the three vectors such that |vec a|=|vec b|=|vec c|=2 and angle between vec a and vec b is (pi)/(3),vec b and vec c and (pi)/(3) and vec a and vec c(pi)/(3). If vec a,vec b and vec c are sides of parallelogram

Prove that vec(a). {(vec(b) + vec(c)) xx (vec(a) + 2vec(b) + 3vec(c))} = [vec(a) vec(b) vec(c)] .

vec(a)_|_vec(b) and vec(c),|vec(a)|=2,|vec(b)|=3,|vec( c )|=4 . The angle between vec(b) and vec( c ) is (2pi)/(3) then |[vec(a) vec(b) vec( c )]| = …………

Three vectors vec(a),vec(b),vec (c ) are such that vec(a) + vec(b) + vec(c ) = vec(0) , if |vec(a)| = 1 |vec(b)| = 4 and |vec(c )|= 2 , then find the value of mu = (vec(a).vec(b) +vec(b).vec(c ) + vec(c ).vec(a))