The equation of the plane which is equidistant from lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}

Question

The equation of the plane which is equidistant from lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x-2}{3}=\frac{y-3}{1}=\frac{z-1}{2}$ is $A x+B y+C z+D=0$ then $|A+B+C+D|=$ $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

Equation of plane which contains the line $\frac{x-2}{3}=\frac{y-3}{1}=\frac{z-1}{2}$ is
$
\begin{aligned}
& (x-2-3(y-3))+\lambda(2(y-3)-(z-1))=0 \
& x+(2 \lambda-3) y-\lambda z+7-5 \lambda=0
\end{aligned}
$
If the plane is parallel to $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$
$
\Rightarrow 1+(2 \lambda-3) 2-3 \lambda=0 \Rightarrow \lambda=5
$
So, equation of plane which contains second line and parallel to first is $x+7 y-5 z-18=0$. \Similarly, equation of plane which contains first line and parallel to second is
$
\begin{aligned}
& (x-1)+7(y-2)-5(z-3)=0 \
& \Rightarrow \quad x+7 y-5 z=0
\end{aligned}
$
Hence equidistant plane is $\frac{(x+7 y-5 z-18)+(x+7 y-5 z)}{2}=0$
$
\Rightarrow \quad x+7 y-5 z-9=0
$
Alternate Solution:
General points on lines are $P(\lambda+1,2 \lambda+2,3 \lambda+3)$ and $Q(3 \mu+2, \mu+3,2 \mu+1)$
Let the required plane be $a x+b y+c z=1$
Then mid point of PQ will always lie on the plane for all values of $\lambda$ and $\mu$
$\Rightarrow a\left(\frac{3 \mu+\lambda+3}{2}ight)+b\left(\frac{2 \lambda+\mu+5}{2}ight)+c\left(\frac{3 \lambda+2 \mu+4}{2}ight)=1$\ $\Rightarrow \quad \mu(3 a+b+2 c)+\lambda(a+2 b+3 c)+3 a+5 b+4 c-2=0 \quad \Rightarrow \quad 3 a+b+2 c=0, a+2 b+3 c=0$\ $3 a+5 b+4 c=2 \Rightarrow a=\frac{1}{9}, b=\frac{7}{9}, c=-\frac{5}{9}$
So, plane is $x+7 y-5 z=9$.