Find the direction cosines of the a vector in the plane of $\overrightarrow{b}=<2,1,3>$ and $\overrightarrow{c}=<-1,2,1>$ and perpendicular to $\overrightarrow{c}$

Let the vector $\overrightarrow{a}=\u27e8x,y,z\u27e9$

Since it is perpendicular to $\overrightarrow{c}$

$-x+2y+z=0$

And

$\left|\begin{array}{ccc}x& y& z\\ 2& 1& 3\\ -1& 2& 1\end{array}\right|$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}-5x-7y+5z=0$

since a, b, c are coplanar

So y=0

Then x=y

So the answer should be $\u27e8\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\u27e9$ , but that is not correct.

What is wrong with my solution?

Let the vector $\overrightarrow{a}=\u27e8x,y,z\u27e9$

Since it is perpendicular to $\overrightarrow{c}$

$-x+2y+z=0$

And

$\left|\begin{array}{ccc}x& y& z\\ 2& 1& 3\\ -1& 2& 1\end{array}\right|$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}-5x-7y+5z=0$

since a, b, c are coplanar

So y=0

Then x=y

So the answer should be $\u27e8\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\u27e9$ , but that is not correct.

What is wrong with my solution?