The plane $ax+by+cz=1$ contains the line $L_2$ and perpendicular to $n$. Determine $a,b,c$.
Sophia Terry 
Let $L_1$ and $L_2$ be two lines.
$L_1:r(t)=\langle1,-2,4\rangle+t\langle4,-3,-1\rangle$
$L_2:s(t)=\langle1,1,0\rangle+t\langle2,3,-2\rangle$
The plane vector $ax+by+cz=1$ contains the line $L_2$ and perpendicular to $n$, where n is the normal to the two lines. Determine $a,b,c$.
My way of doing this was to find the normal to the two lines, which is $\frac{1}{7}\langle3,2,6\rangle$. This is n.
Next I used this formula, $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$, where $A=3$, $B=2$, $C=6$, and the point $\langle1,1,0\rangle$.
I got $3(x-1)+2(y-1)+6(z)=0$, after simplification, got me, $3x+2y+6z=5$
and so, $a=3/5$, $b=2/5$ and $c=6/5$.
My answer is really short though, but I cannot understand.
It says
The equation of the plane perpendicular to $7n=\langle 3,2,6\rangle $ and contains $R$ is $3x+2y+6z=3(1)+2(1)+6(0)=1$.
Hence $a=3/5$, $b=2/5$ and $c=6/5$.
Got not a single idea how it got the $5$ from.
$\endgroup$ 21 Answer
$\begingroup$The equation of the plane perpendicular to $7n=\langle 3,2,6\rangle $ and contains $R$ is $3x+2y+6z=3(1)+2(1)+6(0)=1$.
You say that you're not sure where the 5 came from. The answer is that there's a typo in the section you quoted. $3(1)+2(1)+6(0) = 5$ and there's your 5.
$\endgroup$ 2
