The foot of perpendicular from the origin O to a plane P which meets the co-ordinate axes at the points A,B,C is  $(2,\mathrm{a},4),\mathrm{a}\in \mathrm{N}$. If the volume of the tetrahedron OABC is 144 unit³ , then which of the following points is NOT on P?

Equation of plane is $2(\mathrm{x}-2)+\mathrm{a}(\mathrm{y}-\mathrm{a})+4(\mathrm{z}-4)=0$
$2\mathrm{x}+\mathrm{ay}+4\mathrm{z}=20+{\mathrm{a}}^2 ...\left(1\right)$
$\frac{\mathrm{x}}{\left(\frac{20+{\mathrm{a}}^2}{2}\right)}+\frac{\mathrm{y}}{\left(\frac{20+{\mathrm{a}}^2}{\mathrm{a}}\right)}+\frac{\mathrm{y}}{\left(\frac{20+{\mathrm{a}}^2}{4}\right)}=1$
volume of tetrahedron OABC is $=\frac{1}{6}\frac{\left(20+{\mathrm{a}}^2\right)}{2}\left(\frac{20+{\mathrm{a}}^2}{\mathrm{a}}\right)\cdot \left(\frac{20+{\mathrm{a}}^2}{4}\right)=144$ (given)
${\left(20+{\mathrm{a}}^2\right)}^3=\mathrm{a}.48.144=\mathrm{a}\cdot 12.4\cdot {12}^2=4\mathrm{a}\cdot {12}^3={12}^3\left(4\mathrm{a}\right) ...\left(2\right)$
Eq (2) is satisfied for a = 2
So plane P is  $2\mathrm{x}+2\mathrm{y}+4\mathrm{z}=24\Rightarrow \mathrm{x}+\mathrm{y}+2\mathrm{z}=12 ...\left(3\right)$
Check options (3,0,4) not satisfies (3)