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

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

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

Vector normal to it

$\mathrm{V}=(\mathrm{\lambda }+1)\hat{\mathrm{i}}+(\mathrm{\lambda }-1)\hat{\mathrm{j}}+(2\mathrm{\lambda }-1)\hat{\mathrm{k}}$

Now the vector along the line of intersection of the planes $2\mathrm{x}+3\mathrm{y}+\mathrm{z}-1=0\text{ and }\mathrm{x}+3\mathrm{y}+2\mathrm{z}-2=0$

$\begin{array}{l}\overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& 3& 1\\ 1& 3& 2\end{array}\right|=\left(6-3\right)\hat{\mathrm{i}}-\left(4-1\right)\hat{\mathrm{j}}+\left(6-3\right)\hat{\mathrm{k}}\\ =3(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\end{array}$

As $\overrightarrow{\mathrm{n}}$ is parallel to the plane (1)

$(\mathrm{\lambda }+1)1-1(\mathrm{\lambda }-1)\hat{\mathrm{j}}+\left(2\mathrm{\lambda }-1\right)1=0\Rightarrow \lambda =\frac{-1}{2}$

Hence, the required plane is

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