Let the plane P pass through the intersection of the planes 2x+3y-z=2 and x+2y+3z=6, and b…

Question

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 $d^2$ is equal to:

Accepted Answer

Plane $P$, is passing through intersection of the two planes, so,

$$
\begin{aligned}
& 2 x+3 y-z-2+\lambda(x+2 y+3 z-6)=0 \\
& x(2+\lambda)+y(3+2 \lambda)+z(3 \lambda-1)-2-6 \lambda=0
\end{aligned}
$$

It is perpendicular with plane, $2 x+y-2+1=0$

So, $(2+\lambda) 2+(3+2 \lambda) 1+(3 \lambda-1)(-1)=0$

$\lambda=-8$

So, plane $p_1:-6 x-13 y-25 z+46=0$

Distance of plane $p$ from the point $(-7,1,1)$

$$
\begin{aligned}
& d=\frac{|+42-13-25+46|}{\sqrt{36+169+625}}=\frac{50}{\sqrt{830}} \\
& d^2=\frac{2500}{830}=\frac{250}{83}
\end{aligned}
$$

Solutions · 1