If $\vec a,\vec b,\vec c$ and $\vec d$ are unit vectors such that $(\vec a×\vec b). (\vec c×\vec d) = 1$ and $\vec a.\vec c=\frac{1}{2}$, then

$\vec b,\vec d$ are non-parallel

Since $\vec a,\vec b,\vec c$ and $\vec d$ are unit vectors.

$∴\vec a×\vec b=|\vec a||\vec b|\sin α\hat{n_1}$, 

where $\hat{n_1}$ is a unit vector perpendicular to the plane of $\vec a$ and $\vec b$

and, $\vec c×\vec d=|\vec c||\vec d|\sin β\hat{n_2}$,

where $\hat{n_2}$ is a unit vector perpendicular to the plane of $\vec a$ and $\vec b$

$∴(\vec a×\vec b).(\vec c×\vec d)=1$

$⇒\sin α\,\sin β(\hat{n_1}.\hat{n_2})=1$

$⇒\sin α\,\sin β\cos θ=1$, where θ is the angle between $\hat{n_1}$ and $\hat{n_2}$

$⇒\sin α=1,\sin β=1,\cos θ=1$   $[∵α ∈ [0, π], β∈ [0, π]]$

$⇒α=π/2,β=π/2$ and $θ=0$

Now, $\vec a.\vec c=\frac{1}{2}$

$⇒\cos γ=\frac{1}{2}$, where $γ$ is the angle between $\vec a$ and $\vec c$.

$⇒γ=\frac{π}{3}$

We have, $θ=0$

$∴\hat{n_1}||\hat{n_2}⇒\vec a×\vec b||\vec c×\vec d⇒\vec a,\vec b,\vec c$ are non-coplanar

Thus, we have $\vec a⊥\vec b, \vec c⊥\vec d$ and angle between $\vec a$ and $\vec c$ is $π/3$.

So, we have the following possibilities.

In both the cases, we find that $\vec b$ is non-parallel to $\vec d$.

Hence, option (3) is correct.