A plane is parallel to two lines whose direction ratios are \(1,0,-1\) and \(-1,1,0\) and it contains the point \((1,1,1)\). If it cuts the co-ordinate axes at \(\mathrm{A}, \mathrm{B}, \mathrm{C}\), then the volume of the tetrahedron \(\mathrm{OABC}\) (in cubic units) is

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
\(
\begin{aligned}
& a-c=0 \text { and }-a+b=0 \\
& \Rightarrow a=b=c
\end{aligned}
\)
\(\therefore\) From (i) and (ii), we get
\(
\begin{aligned}
& x-1+y-1+z-1=0 \\
& \therefore x+y+z=3 \Rightarrow \frac{x}{3}+\frac{y}{3}+\frac{z}{3}=1
\end{aligned}
\)
\(\therefore\) Co-ordinates of A, B, C are \((3,0,0),(0,3,0)\) and \((0,0,3)\) respectively.
\(\therefore\) Volume of tetrahedron \(\mathrm{OABC}\)
\(
\begin{aligned}
& =\frac{1}{6}\left|\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right| \\
& =\frac{1}{6} \times 27 \\
& =\frac{9}{2} \text { cu. units }
\end{aligned}
\)