The vectors $\vec a,\vec b$ and $\vec c$ are of the same length and taken pairwise, they from equal angles. If $\vec a=\hat i+\hat j$ and $\vec b=\hat j+\hat k$ then $\vec c$ is equal to:

$\hat i+\hat k$

$\vec a.\vec b=|\vec a||\vec b|cosθ⇒1=(\sqrt{2})(\sqrt{2})cosθ⇒θ=\frac{\pi}{3}$

Let $\vec c=x\hat i+y\hat j+z\hat k$

$\vec a.\vec c=\vec b.\vec c=\vec a.\vec b$ as the angel is same and $|\vec a|=|\vec b|=|\vec c|$

$⇒x+y=y+z=1$  $⇒x=z$

Also $|c^2|=2⇒x^2+y^2+x^2=2⇒2x^2+y^2=2$

Solving x + y = 1 and $2x^2+y^2=2$, we get;

$⇒3x^2-2x-1=0⇒(3x+1)(x-1)=0⇒x=1,\frac{-1}{3}$

$⇒\vec c=\hat i+\hat k$ or $\frac{-1}{3}\hat i+\frac{4}{3}\hat j-\frac{1}{3}\hat k$