Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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