# Find the equation of the plane perpendicular to the plane $P$ and containing line $L_1$.

Consider a plane $P$ passing through $A(\lambda,3,\mu)$,$B(-1,3,2)$ and $C(7,5,10)$ and a straight line $L$ with positive direction cosines passing through $A$,bisecting $BC$ and makes equal angles with the coordinate axes.Let $L_1$ be a line parallel to $L$ and passing through origin.Find the equation of the plane perpendicular to the plane $P$ and containing line $L_1$.

Let the direction cosines of the line $L$ are $l,m,n$.Since the line $L$ has positive direction cosines and it makes equal angles with the coordinate axes. So $l=m=n$.
We know,$l^2+m^2+n^2=1$
$\Rightarrow l=m=n=\frac{1}{\sqrt{3}}$
Since the line $L$ bisects the $BC$,so it passes through $(3,4,6)$
Symmetric form of the line $L$ is $\frac{x-3}{\frac{1}{\sqrt3}}=\frac{y-4}{\frac{1}{\sqrt3}}=\frac{z-6}{\frac{1}{\sqrt3}}$
Since the line $L$ also passes through $A(\lambda,3,\mu)$
So $\frac{\lambda-3}{\frac{1}{\sqrt3}}=\frac{3-4}{\frac{1}{\sqrt3}}=\frac{\mu-6}{\frac{1}{\sqrt3}}$
$\lambda=2,\mu=5$
Equation of the line $L_1$ is $\frac{x}{1}=\frac{y}{1}=\frac{z}{1}$
As the plane $P$ passes through $A(2,3,5),B(-1,3,2),C(7,5,10)$.I found its equation as $x-z+3=0$

But i cannot find the equation of the required plane which is perpendicular to the plane $P$ and contains line $L_1.$Please help me.

This plane must contain the point $D(1,1,1)$, say, on $L_1$ and the point $E$ such that $\overrightarrow{OE}=\overrightarrow{AB}\times\overrightarrow{AC}$.
Well so we need to find the equation of the plane, say $Q$, which is perpendicular to $P$ and contains line $L_1$.
Since $Q$ is perpendicular to $P$, a line perpendicular to plane $P$, say $N$, will be parallel to the required plane $Q$. The line $N$ can be found by taking cross product of $\vec{AB}$ and $\vec{CB}$. And since $L_1$ lies on plane $Q$, a cross-product of $N$ and $L_1$ will give us the perpendicular vector to plane $Q$.
Now since the line $L_1$, which passes through origin, lies on plane $Q$, origin also lies on plane $Q$.
And now we have the perpendicular to the plane $Q$ and a point on that plane, it is easy to find the equation of the required plane $Q$.