The correct option is
D All
→a,→b,→cGiven
→a.→b=→b.→c=→c.→a=0
And
|→a|=|→b|=|→c|=k (say)
→a.(→a+→b+→c)=|→a||→a+→b+→c|cosα
Where α is the angle between →a and →a+→b+→c
→a.→a+→a.→b+→a.→c=k|→a+→b+→c|cosα
|→a|2+→a.→b+→a.→c=k|→a+→b+→c|cosα
k2=k|→a+→b+→c|cosα
cosα=k|→a+→b+→c| ----- (i)
→b.(→a+→b+→c)=|→b||→a+→b+→c|cosβ
Where β is the angle between →b and (→a+→b+→c)
→b.→a+→b.→b+→b.→c=k|→a+→b+→c|cosβ
|→b|2=k|→a+→b+→c|cosβ
k2=k|→a+→b+→c|cosβ
cosβ=k|→a+→b+→c| ----- (ii)
Similarly,
→c.(→a+→b+→c)=|→c||→a+→b+→c|cosγ
Where γ is the angle between →c and (→a+→b+→c)
→c.→a+→c.→b+→c.→c=k|→a+→b+→c|cosγ
|→c|2=k|→a+→b+→c|cosγ
k2=k|→a+→b+→c|cosγ
cosγ=k|→a+→b+→c| ----- (iii)
From (i),(ii), and (iii)
cosα=cosβ=cosγ
Hence →a+→b+→c is equally inclined to all →a,→b and →c
Hence, Option C is the correct answer.