The vector along the line of intersection will be $\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a& b& 0\\ a& b& c\end{array}\right|$

$=bc\hat{i}-ac\hat{j}$

i.e. direction ratios of line are $b,-a,0$

Direction ratio of normal to the plane $y-z+2=0$ are $0,1,-1$

Since the line makes $30°$ with the plane, so it will make $60°$ with its normal.

i.e. $\cos 60°=\left|\frac{-a}{\sqrt{2}\sqrt{b^2+a^2}}\right|$

$\Rightarrow \left|\frac{a}{\sqrt{2}\sqrt{a^2+b^2}}\right|=\frac{1}{2}$

$\Rightarrow b=\pm a$

So, direction ratios of the line can be $\left(\pm a,-a,0\right)$

$\therefore$  direction cosines can be $\pm \left(\frac{\pm 1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0\right)$