Let $\vec α = a\hat i +b\hat j+c\hat k, \vec β =b\hat i+c\hat j+a\hat k$ and $\vec γ=c\hat i+a\hat j+b\hat k$ be three coplanar vectors with $a≠b$, and $\vec v=\hat i+\hat j+\hat k$. Then, $\vec v$ is perpendicular to

all of these

It is given that $\vec α, \vec β$ and $\vec γ$ are coplanar vectors.

$∴[\vec α\,\,\vec β\,\,\vec γ]=0$

$⇒\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}=0$

$⇒3abc-a^3-b^3-c^3 =0$

$⇒a^3+b^3+ c^3 -3 abc = 0$

$⇒(a+b+c) (a^2 + b^2 + c^2-ab-bc-ca) = 0$

$⇒a+b+c=0$   $[∵ a^2+b^2 + c^2-ab-bc - ca = 0]$

$⇒\vec v.\vec α=\vec v.\vec β=\vec v.\vec γ=0$

$⇒\vec v$ perpendicular to $\vec α, \vec β$ and $\vec γ$.