It is given that
`|vec(a)|=|vec(b)|=|vec(c)|=a (say) ...(i)
Since `vec(a), vec(b)` and `vec(c)` are mutually perpendicular vectors, we have
`vec(a).vec(b)=vec(b).vec(c)=vec(c).vec(a)=0` ...(ii)
Now, `|vec(a)+vec(b)+vec(c)|^(2)=(vec(a)+vec(b)+vec(c)).(vec(a)+vec(b)+vec(c))`
`=vec(a).vec(a)+vec(b).vec(b)+vec(c).vec(c)+2 (vec(a).vec(b)+vec(b).vec(c)+vec(c).vec(a))`
`=|vec(a)|^(2)+|vec(b)|^(2)+|vec(c)|^(2)` [using (ii)]
`= 3 a^(2)` [using (i)].
`:. |vec(a)+vec(b)+vec(c)|=sqrt(3)a`. ...(iii)
Let `(vec(a)+vec(b)+vec(c))` make angles `alpha, beta` and `gamma` with `vec(a), vec(b)` and `vec(c)` respectively.
Then `(vec(a)+vec(b)+vec(c)).vec(a)=|vec(a)+vec(b)+vec(c)||vec(a)| cos alpha`
`= (sqrt(3)axxaxxcos alpha)=sqrt(3) a^(2) cos alpha`
`implies (vec(a).vec(a)+vec(b).vec(a)+vec(c).vec(a))=sqrt(3)a^(2) cos alpha`
`implies |vec(a)|^(2)=sqrt(3) a^(2) cos alpha implies a^(2) = sqrt(3) a^(2) cos alpha`
`implies cos alpha = 1/sqrt(3) implies alpha = cos^(-1) (1/sqrt(3))`.
Similarly, `beta = cos^(-1) (1/sqrt(3))` and `gamma=cos^(-1) (1/sqrt(3))`.
`:. alpha=beta=gamma = cos^(-1) (1/sqrt(3))`.
Hence, `(vec(a)+vec(b)+vec(c))` is equally inclined to `vec(a), vec(b)` and `vec(c)` and the required angle is `cos^(-1) (1/sqrt(3))`.
