Equation of the plane passing through $(1,1,1)$ is given as
$$
\mathrm{a}(x-1)+\mathrm{b}(y-1)+\mathrm{c}(\mathrm{z}-1)=0
$$
As the plane is parallel to the lines having direction ratios $1,0,-1$ and $-1,1,0$, we get
$$
\begin{aligned}
& a-c=0 \text { and }-a+b=0 \\
& \Rightarrow a=b=c
\end{aligned}
$$
$\therefore \quad$ From (i) and (ii), we get
$$
\begin{aligned}
& x-1+y-1+z-1=0 \\
& \therefore \quad x+y+z=3 \quad \Rightarrow \frac{x}{3}+\frac{y}{3}+\frac{z}{3}=1
\end{aligned}
$$
$\therefore \quad$ Co-ordinates of A, B, C are $(3,0,0),(0,3,0)$ and $(0,0,3)$ respectively.
$\therefore \quad$ Volume of tetrahedron $\mathrm{OABC}$
$$
\begin{aligned}
& =\frac{1}{6}\left|\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right| \\
& =\frac{1}{6} \times 27 \\
& =\frac{9}{2} \text { cu. units }
\end{aligned}
$$