A plane P contains the line of intersection of the plane r (i+j+k)=6 and r (2i+3j+4k)=-5.

Question

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 $P$ passes through the point $(0,2,-2)$, then the square of distance of the point $(12,12,18)$ from the plane $P$ is :

Accepted Answer

Given plane $\vec{r}\cdot (\hat{i}+\hat{j}+\hat{k})=6$ and

$\vec{r}\cdot (2\hat{i}+3\hat{j}+4\hat{k})=-5$

Equation of plane passing through both plane

$\begin{gathered} {P}_{1}	o (x\hat{i}+y\hat{j}+2\hat{k})(\hat{i}+\hat{j}+\hat{k})=6 \ {P}_{1}=x+y+z=6 \ {P}_{2}	o (x\hat{i}+y\hat{j}+z\hat{k})\cdot (2\hat{i}+3\hat{j}+4\hat{k})=-5 \ {P}_{2}	o =2x+3y+4z=-5 \end{gathered}$

$\begin{gathered} {P}_{1}+\lambda {P}_{2}=0 \ \Rightarrow (x+y+z-6)+\lambda (2x+3y+4z+5)=0 \end{gathered}$

passes through $(0,2,-2)$

$\begin{gathered} \Rightarrow (0+2-2-6)+\lambda (2	imes 0+3	imes 2+4	imes (-2)+5)=0 \ \Rightarrow \lambda =2 \end{gathered}$

Equation of plane

$5x+7y+9z+4=0$

$	herefore $ Distance of the point $(12,12,18)$ from the plane

$\begin{gathered} d =\frac{5	imes 12+7	imes 12+9	imes 18+4}{\sqrt{{5}^{2}+{7}^{2}+{9}^{2}}} \ \ =\frac{60+84+162+4}{\sqrt{155}}=\frac{310}{\sqrt{155}} \ \ 	herefore {d}^{2} =\frac{310	imes 310}{155}=620 \end{gathered}$

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