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A plane is parallel to two lines 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 A, B, C, then the volume of the tetrahedron $\mathrm{OABC}$ is cu. units.
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The correct answer is:
$\frac{9}{2}$
Direction ration of normal to the plane can be obtained by
$\begin{aligned} & \frac{\mathrm{a}}{0 \times 0-1 \times(-1)}=\frac{\mathrm{b}}{(-1) \times(-1)-1 \times 0}=\frac{\mathrm{c}}{1 \times 1-(-1) \times 0} \\ & \Rightarrow\langle\mathrm{a}, \mathrm{b}, \mathrm{c}\rangle \equiv\langle 1,1,1\rangle\end{aligned}$
Hence equation of the plane $1(x-1)+1(y-1)+1(z-1)=0$
$\begin{aligned} & \Rightarrow x+y+z=3 \\ & \Rightarrow \frac{x}{3}+\frac{y}{3}+\frac{z}{3}=1\end{aligned}$
$\Rightarrow$ The plane intersects the co-ordinate axes at $(3,0,0),(0,3,0)$ and $(0,0,3)$
Hence, volume of tetrahedron OABC is
$\begin{aligned} & \frac{\mathrm{a}}{0 \times 0-1 \times(-1)}=\frac{\mathrm{b}}{(-1) \times(-1)-1 \times 0}=\frac{\mathrm{c}}{1 \times 1-(-1) \times 0} \\ & \Rightarrow\langle\mathrm{a}, \mathrm{b}, \mathrm{c}\rangle \equiv\langle 1,1,1\rangle\end{aligned}$
Hence equation of the plane $1(x-1)+1(y-1)+1(z-1)=0$
$\begin{aligned} & \Rightarrow x+y+z=3 \\ & \Rightarrow \frac{x}{3}+\frac{y}{3}+\frac{z}{3}=1\end{aligned}$
$\Rightarrow$ The plane intersects the co-ordinate axes at $(3,0,0),(0,3,0)$ and $(0,0,3)$
Hence, volume of tetrahedron OABC is
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