A plane P contains the line of intersection of the plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5$. If $\mathrm{P}$ passes through the point $(0,2,-2)$, then the square of distance of the point $(12,12,18)$ from the plane $\mathrm{P}$ is :

Given plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$ and <br/><br/>$\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5$ <br/><br/>Equation of plane passing through both plane <br/><br/>$$ \begin{aligned} & \mathrm{P}_1 \rightarrow(x \hat{i}+y \hat{j}+2 \hat{k})(\hat{i}+\hat{j}+\hat{k})=6 \\\\ & \mathrm{P}_1=x+y+z=6 \\\\ & \mathrm{P}_2 \rightarrow(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5 \\\\ & \mathrm{P}_2 \rightarrow=2 x+3 y+4 z=-5 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \mathrm{P}_1+\lambda \mathrm{P}_2=0 \\\\ & \Rightarrow(x+y+z-6)+\lambda(2 x+3 y+4 z+5)=0 \end{aligned} $$ <br/><br/>passes through $(0,2,-2)$ <br/><br/>$$ \begin{aligned} & \Rightarrow(0+2-2-6)+\lambda(2 \times 0+3 \times 2+4 \times(-2)+5)=0 \\\\ & \Rightarrow \lambda=2 \end{aligned} $$ <br/><br/>Equation of plane <br/><br/>$5 x+7 y+9 z+4=0$ <br/><br/>$\therefore$ Distance of the point $(12,12,18)$ from the plane <br/><br/>$$ \begin{aligned} d & =\frac{5 \times 12+7 \times 12+9 \times 18+4}{\sqrt{5^2+7^2+9^2}} \\\\ & =\frac{60+84+162+4}{\sqrt{155}}=\frac{310}{\sqrt{155}} \\\\ \therefore d^2 & =\frac{310 \times 310}{155}=620 \end{aligned} $$