The foot of perpendicular from the origin $\mathrm{O}$ to a plane $\mathrm{P}$ which meets the co-ordinate axes at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ is $(2, \mathrm{a}, 4), \mathrm{a} \in \mathrm{N}$. If the volume of the tetrahedron $\mathrm{OABC}$ is 144 unit$^{3}$, then which of the following points is NOT on P ?

Equation of Plane: <br/><br/>$$ \begin{aligned} & (2 \hat{i}+a \hat{j}+4 \hat{\mathrm{k}}) \cdot[(\mathrm{x}-2) \hat{\mathrm{i}}+(\mathrm{y}-\mathrm{a}) \hat{\mathrm{j}}+(\mathrm{z}-4) \hat{\mathrm{k}}]=0 \\\\ & \Rightarrow 2 \mathrm{x}+\mathrm{ay}+4 \mathrm{z}=20+\mathrm{a}^{2} \\\\ & \Rightarrow \mathrm{A} \equiv\left(\frac{20+\mathrm{a}^{2}}{2}, 0,0\right) \\\\ & \mathrm{B} \equiv\left(0, \frac{20+\mathrm{a}^{2}}{\mathrm{a}}, 0\right) \\\\ & \mathrm{C} \equiv\left(0,0, \frac{20+\mathrm{a}^{2}}{4}\right) \\\\ & \Rightarrow \text { Volume of tetrahedron } \\\\ & =\frac{1}{6}\left[ {\matrix{ {\overrightarrow a } & {\overrightarrow b } & {\overrightarrow c } \cr } } \right] \\\\ & =\frac{1}{6} \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}) \\\\ & \Rightarrow \frac{1}{6}\left(\frac{20+\mathrm{a}^{2}}{2}\right) \cdot\left(\frac{20+\mathrm{a}^{2}}{\mathrm{a}}\right) \cdot\left(\frac{20+\mathrm{a}^{2}}{4}\right)=144 \\\\ & \Rightarrow\left(20+\mathrm{a}^{2}\right)^{3}=144 \times 48 \times \mathrm{a} \\\\ & \Rightarrow \mathrm{a}=2 \\\\ & \Rightarrow \text { Equation of plane is } 2 \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}=24 \\\\ & \text { Or } \mathrm{x}+\mathrm{y}+2 \mathrm{z}=12 \\\\ & \Rightarrow(3,0,4) \quad \text { Not } \text { lies on } \text { the } \\\\ & \mathrm{x}+\mathrm{y}+2 \mathrm{z}=12 \end{aligned} $$