The equation of the plane passing

Equation of planes are

$\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})-1=0 \Rightarrow \mathrm{x}+\mathrm{y}+\mathrm{z}-1=0$

and $\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})+4=0 \Rightarrow 2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}+4=$ 0 equation of planes through line of intersection of these planes is :-

$(x+y+z-1)+\lambda(2 x+3 y-z+4)=0$

$\Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z-1+4 \lambda=0$

But this plane is parallel to $x$-axis whose direction are $(1,0,0)$

$\therefore(1+2 \lambda) 1+(1+3 \lambda) 0+(1-\lambda) 0=0$

$\lambda=-\frac{1}{2}$

$\therefore$ Required plane is

$0 x+\left(1-\frac{3}{2}\right) y+\left(1+\frac{1}{2}\right) z-1+4\left(\frac{-1}{2}\right)=0$

$\Rightarrow \frac{-\mathrm{y}}{2}+\frac{3}{2} \mathrm{z}-3=0$

$\Rightarrow \mathrm{y}-3 \mathrm{z}+6=0$

$\Rightarrow \overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{j}}-3 \hat{\mathrm{k}})+6=0$ Ans.