Let $\overrightarrow{\mathrm{a}}=(2,1,-1), \overrightarrow{\mathrm{b}}=(1,-1,0), \overrightarrow{\mathrm{c}}=(5,-1,1)$
$\therefore \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}=(-2,1,-2)$
Let $\overrightarrow{\mathrm{n}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$ be the unit vector which is $\|$ to $(-2$
$1,-2$ ) in the opposite direction.
$\therefore \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=1$ and $\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \mathrm{x} & \mathrm{y} & \mathrm{z} \\ -2 & 1 & -2\end{array}\right|=0$
$\Rightarrow x=-2 y, y=y, z=-2 y$
$x^{2}+y^{2}+z^{2}=1 \Rightarrow 4 y^{2}+y^{2}+4 y^{2}=1 \Rightarrow y=\pm \frac{1}{3}$
Hence, the Required vector
$\hat{n}=\frac{2}{3} \hat{i}-\frac{\hat{j}}{3}+\frac{2}{3} \hat{k}$