A plane is parallel to two lines whose direction ratios are \(2,0,-2\) and \(-2,2,0\) and it contains the point \((2,2,2)\). If it cuts co-ordinate axes at \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) then the volume of tetrahedron \(\mathrm{OABC}\) (in cubic units) is

The equation of plane passing through \((2,2,2)\)
is \(\mathrm{a}(x-2)+\mathrm{b}(y-2)+\mathrm{c}(\mathrm{z}-2)=0\)
Also, \(2 \mathrm{a}-2 \mathrm{c}=0\)
\(-2 a+2 b=0\)
\(\Rightarrow \mathrm{a}=\mathrm{b}=\mathrm{c}\)
\(\begin{aligned} & x+y+z-6=0 \\ & \Rightarrow \frac{x}{6}+\frac{y}{6}+\frac{z}{6}=1\end{aligned}\)
It cuts the co-ordinate axes at \(\mathrm{A}(6,0,0)\), \(\mathrm{B}(0,6,0)\) and \(\mathrm{C}(0,0,6)\)
\(\therefore \quad \overline{\mathrm{a}}=6 \hat{\mathrm{i}}, \overline{\mathrm{b}}=6 \hat{\mathrm{j}}, \overline{\mathrm{c}}=6 \hat{\mathrm{k}}\)
\(\therefore \quad\) Volume of tetrahedron \(=\frac{1}{6}\left[\begin{array}{lll}-\bar{a} & \bar{b} & \bar{c}\end{array}\right]\)
\(\begin{aligned} & =\frac{1}{6}\left|\begin{array}{lll}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right| \\ & =36 \text { cubic units }\end{aligned}\)