Let the equation of the plane containing line $\mathrm{x}-\mathrm{y}-\mathrm{z}-4=0=\mathrm{x}+\mathrm{y}+2\mathrm{z}-4$ and parallel to the line of intersection of the planes $\mathrm{y}+3\mathrm{y}+\mathrm{z}=1$ and $\mathrm{x}+ 3\mathrm{y}+27=2$be $\mathrm{x}+\mathrm{Ay}+\mathrm{Bz}+\mathrm{C}=0$ Then find the value of $|\mathrm{A}+\mathrm{B}+\mathrm{C}|=$

A plane containing the line of intersection of the given planes is 

$\begin{array}{l}\mathrm{x}-\mathrm{y}-\mathrm{z}-4+\mathrm{\lambda }(\mathrm{x}+\mathrm{y}+2\mathrm{z}-4)=0\\ \text{ i.e. }(\mathrm{\lambda }+1)\mathrm{x}+(\mathrm{\lambda }-1)\mathrm{y}+(2\mathrm{\lambda }-1)\mathrm{z}-4(\mathrm{\lambda }+1)=0\end{array}$

vector normal to it 

$\mathrm{V}=(\mathrm{\lambda }+1)\widehat{\mathrm{i}}+(\mathrm{\lambda }-1)\widehat{\mathrm{j}}+(2\mathrm{\lambda }-1)\widehat{\mathrm{k}}\mathrm{S}----\left(\mathrm{i}\right)$

Now the vector along the line of intersection of the planes 2x + 3y + z - 1 = 0 and x + 3y + 2z -2 = 0 is given by

$\overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 2& 3& 1\\ 1& 3& 2\end{array}\right|=3(\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}})$

As $\overrightarrow{\mathrm{n}}$is parallel to the plane (i), we have 

$\begin{array}{l}\overrightarrow{\mathrm{n}}\cdot \overrightarrow{\mathrm{V}}=0\\ (\mathrm{\lambda }+1)-(\mathrm{\lambda }-1)+(2\mathrm{\lambda }-1)=0\\ 2+2\mathrm{\lambda }-1=0\text{ or }\mathrm{\lambda }=\frac{-1}{2}\end{array}$

Hence, the required plane is
$\begin{array}{l}\frac{\mathrm{x}}{2}-\frac{3\mathrm{y}}{2}-2\mathrm{z}-2=0\\ \mathrm{x}-3\mathrm{y}-4\mathrm{z}-4=0\\ \mathrm{Hence},|\mathrm{A}+\mathrm{B}+\mathrm{C}|=11\end{array}$