If the equation of the plane containing the line

$x+2 y+3 z-4=0=2 x+y-z+5$ and perpendicular to the plane

$\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})$

is $a x+b y+c z=4$, then $(a-b+c)$ is equal to :

Equation of plane $\mathrm{P}$ containing the given lines is <br/><br/>$$ \begin{aligned} & (x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0 \\\\ & \Rightarrow(1+2 \lambda) x+(2+\lambda) y+(3-\lambda) z+(-4+5 \lambda)=0 \end{aligned} $$ <br/><br/>Now, plane $\mathrm{P}$ is perpendicular to plane $\mathrm{P}^{\prime}$ <br/><br/>$$ \vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k}) $$ <br/><br/>So, normal to plane $\mathrm{P}^{\prime}$ is <br/><br/>$$ \begin{aligned} & \vec{n}=(\hat{i}+\hat{j}+\hat{k}) \times(\hat{i}-2 \hat{j}+3 \hat{k}) \\\\ & \Rightarrow \vec{n}=5 \hat{i}-2 \hat{j}-3 \hat{k} \end{aligned} $$ <br/><br/>$\therefore \mathrm{P}$ and $\mathrm{P}^{\prime}$ are perpendicular <br/><br/>$$ \begin{aligned} & \therefore 5(1+2 \lambda)-2(2+\lambda)-3(3-\lambda)=0 \\\\ & \Rightarrow 5+10 \lambda-4-2 \lambda-9+3 \lambda=0 \\\\ & \Rightarrow 11 \lambda=8 \Rightarrow \lambda=\frac{8}{11} \end{aligned} $$ <br/><br/>$$ \begin{array}{r} \therefore P:\left(1+\frac{16}{11}\right) x+\left(2+\frac{8}{11}\right) y+\left(3-\frac{8}{11}\right) z+\left(5 \times \frac{8}{11}-4\right) =0 \end{array} $$ <br/><br/>i.e., $27 x+30 y+25 z=4$ <br/><br/>which is same as $a x+b y+c z=4$ <br/><br/>$$ \begin{aligned} & \therefore a=27, b=30 \text { and } c=25 \\\\ & \Rightarrow a-b+c=27-30+25=22 \end{aligned} $$