The direction cosines of the line $x-y + 2z = 5, $   $3x + y + z = 6 $ are

$\frac{-3}{5\sqrt{2}}, \frac{5}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}$

Vectors normal tp given planes are $\vec{n_1}=\hat{i}-\hat{j} + 2 \hat{k}$ and $\vec{n_2}= 3 \hat{i}+\hat{j} +  \hat{k}$.

So, their line of intersection is parallel to the vector

$ \vec{n}= \vec{n_1}×\vec{n_2}= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\1 & -1 & 2\\3 & 1 & 1\end{vmatrix}= -3 \hat{i}+5\hat{j} +  4\hat{k}$

$⇒ \vec{n} = \frac{-3}{5\sqrt{2}}\hat{i}+\frac{5}{5\sqrt{2}}\hat{j} +  \frac{4}{5\sqrt{2}}\hat{k}$

hence, direction cosines of the line are $\frac{-3}{5\sqrt{2}}, \frac{5}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}$