Let the plane P pass through the intersection of the planes $2x+3y-z=2$ and $x+2y+3z=6$, and be perpendicular to the plane $2x+y-z+1=0$. If d is the distance of P from the point ($-$7, 1, 1), then $\mathrm{d^{2}}$ is equal to :
Solution
Plane $P$, is passing through intersection of the two planes, so,
<br/><br/>$$
\begin{aligned}
& 2 x+3 y-z-2+\lambda(x+2 y+3 z-6)=0 \\\\
& x(2+\lambda)+y(3+2 \lambda)+z(3 \lambda-1)-2-6 \lambda=0
\end{aligned}
$$
<br/><br/>It is perpendicular with plane, $2 x+y-2+1=0$
<br/><br/>So, $(2+\lambda) 2+(3+2 \lambda) 1+(3 \lambda-1)(-1)=0$
<br/><br/>$\lambda=-8$
<br/><br/>So, plane $p_1:-6 x-13 y-25 z+46=0$
<br/><br/>Distance of plane $p$ from the point $(-7,1,1)$
<br/><br/>$$
\begin{aligned}
& d=\frac{|+42-13-25+46|}{\sqrt{36+169+625}}=\frac{50}{\sqrt{830}} \\\\
& d^2=\frac{2500}{830}=\frac{250}{83}
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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