We are given $\vec a = x\hat i + y\hat j + z\hat k, \qquad \hat b = \hat j.$ We need to find $\vec c$ such that $\vec a, \vec c, \hat b$ form a right-handed system.

For three vectors $\vec a, \vec c, \vec b$ to form a right-handed system, the standard relation is $\vec a \times \vec c = \vec b.$ Here, $\vec a \times \vec c = \hat j.$

Let $\vec c = p\hat i + q\hat j + r\hat k.$ Then

Since this must equal $\hat j$, we require $yr-zq=0,$ $-(xr-zp)=1,$ $xq-yp=0.$

Now test the options.

Option A: $\vec c = z\hat i - x\hat k.$ So $p=z, q=0, r=-x$.

Compute:

In such questions, the intended meaning is usually that $\vec c$ is perpendicular to both $\vec a$ and $\hat b$, and chosen so that $\vec a,\vec c,\hat b$ are right-handed. Hence $\vec c = \hat b \times \vec a.$

Since for a right-handed triplet $(\vec a,\vec c,\hat b)$, cyclic order gives $\vec c = \hat b \times \vec a.$

Now compute: $\vec c = \hat j \times (x\hat i + y\hat j + z\hat k).$

Using $\hat j \times \hat i = -\hat k, \qquad \hat j \times \hat j = \vec 0, \qquad \hat j \times \hat k = \hat i,$ we get $\vec c = x(\hat j \times \hat i) + y(\hat j \times \hat j) + z(\hat j \times \hat k)$ $= x(-\hat k) + 0 + z\hat i$ $= z\hat i - x\hat k.$

Therefore, $\boxed{\vec c = z\hat i - x\hat k}.$ This is Option A.