If the vectors $\vec c, \vec a = x\hat i + y\hat j +z\hat k$ and $\vec b = \hat j$ are such that $\vec a, \vec c$ and $\vec b$ form a right handed system, then $\vec c$, is

$z\hat i-x\hat k$

Since $\vec a,\vec c,\vec b$ form a right handed system.

$∴\vec a×\vec c=\vec b,\vec c×\vec a=\vec a$ and $\vec b×\vec a=\vec c$

Now,

$\vec c=\vec b×\vec a⇒\vec c=\begin{vmatrix}\hat i&\hat j&\hat k\\0&1&0\\x&y&z\end{vmatrix}=z\hat i-x\hat k$