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

$\hat i +\hat k, \frac{1}{3}(-\hat i + 4\hat j-\hat k)$

Let $c=x\hat i+y\hat j+z\hat k$. Then,

$|\vec a|=|\vec b|=|\vec c| ⇒ x^2 + y^2+z^2=2$   ...(i)

It is given that the angles between the vectors taken in pairs are equal, say, θ.

$∴\cos θ=\frac{\vec a.\vec b}{|\vec a||\vec b|}=\frac{0+1+0}{\sqrt{2}\sqrt{2}}=\frac{1}{2}$

$⇒\frac{\vec a.\vec c}{|\vec a||\vec c|}=\frac{1}{2}$ and $\frac{\vec b.\vec c}{|\vec b||\vec c|}=\frac{1}{2}$

$⇒\frac{x+y}{\sqrt{2}\sqrt{2}}=\frac{1}{2}$ and $\frac{y+z}{\sqrt{2}\sqrt{2}}=\frac{1}{2}$

$⇒x+y=1$ and $y + z=1$

$⇒y=1-x$ and $z=1-y=1-(1-x) = x$

$∴x^2 + y^2+z^2 = 2$

$⇒x^2 +(1-x)^2 + x^2 = 2$

$⇒3x^2-2x-1=0⇒(3x+1)(x-1)=0⇒x=1,-1/3$.

Now, $y=1-x= y = 0$ for $x=1$ and $y = 4/3$ for $x=-1/3$

Hence, $\vec c=\hat i+0\hat j+\hat k$ or, $\vec c=-\frac{1}{3}\hat i +\frac{4}{3}\hat j-\frac{1}{3}\hat k$