I know how to solve a question like this when we have to find a vector equally inclined to only Two vectors-
let these two vectors are (A) and (B). Then you find A^ and B^ {I mean unit vectors} and simply add them to find the diagonal of the rhombus made by these unit vectors. This new diagonal vector is the required vector.
Now I applied this approach for my problem with the three vectors but it does not works. Please help.
On
Suppsoe your unit vectors are $\vec a$, $\vec b$ and $\vec c$. The cosine of the angle between any two vectors $\vec x$ and $\vec y$ is their dot product divided by the product of their magnitudes i.e.
$\displaystyle \cos \theta = \frac{\vec x . \vec y}{| \vec x || \vec y |}$
In the 2D case, if $\vec x = \vec a + \vec b$, then we have
$\vec x . \vec a = \vec a . \vec a + \vec b . \vec a = 1 + \vec a . \vec b \\ \vec x . \vec b= \vec a . \vec b+ \vec b. \vec b = 1 + \vec a . \vec b \\ \Rightarrow \vec x . \vec a = \vec x . \vec b$
and so the angle between $\vec x$ and $\vec a$ is equal to the angle between $\vec x$ and $\vec b$ (strictly speaking, it is equal in magnitude but opposite in direction).
However, in the 3D case, if we take $\vec x = \vec a + \vec b + \vec c$ then we have
$\vec x . \vec a = \vec a . \vec a + \vec b . \vec a + \vec c . \vec a = 1 + \vec a . \vec b + \vec a . \vec c\\ \vec x . \vec b= \vec a . \vec b+ \vec b. \vec b + \vec c. \vec b = 1 + \vec a . \vec b + \vec b . \vec c\\ \vec x . \vec c= \vec a . \vec c+ \vec b. \vec c + \vec c. \vec c = 1 + \vec a . \vec c + \vec b . \vec c\\$
and these three values are, in general, not equal. You have to find some other linear combination of $\vec a$, $\vec b$ and $\vec c$ which will make $\vec x . \vec a = \vec x . \vec b = \vec x . \vec c$.
You have to take the vector as $$x\hat{i}+y\hat{j}+z\hat{k}$$ and then you have to take dot product of this vector with other three unit vectors, then three variables three equations you can solve it.