The equation of the plane passing through the line of intersection of the planes $x+2y+3z-2=0\:\:and\:\:x-y+z-3=0$ and which is at a distance of $\large\frac{2}{\sqrt 3}$ units from $(3,1,-1)$. is ?

Toolbox:
• General equation of all the planes passing through the line of intersection of te planes $a_1x+b_1y+c_1z+d_1=0\:\:and\:\:a_2x+b_2y+c_2z+d_2=0$ is given by $(a_1x+b_1y+c_1z+d_1)+\lambda(a_1x+b_1y+c_1z+d_1)=0$
• $\perp$ distance of the plane $ax+by+cz+d=0$ from $(x_1,y_1,z_1)$ is $\bigg|\large\frac{ax_1+by_1+cz_1+d}{\sqrt {a^2+b^2+c^2}}\bigg|$
Given two planes are $x+2y+3z-2=0$..........(i) and $x-y+z-3=0$.......(ii)
General eqn. of the plane passing through the line of intersection of (i) and (ii) is
$(x+2y+3z-2)+\lambda(x-y+z-3)=0$
$\Rightarrow\:x(1+\lambda)+y(2-\lambda)+z(3+\lambda)+(-2-3\lambda)=0$....(iii)
Given that the plane (iii) is at a distance of $\large\frac{2}{\sqrt 3}$ from $(3,1,-1)$.
$\therefore\:\bigg|\large\frac{3(1+\lambda)+(2-\lambda)-(3+\lambda)+(-2-3\lambda)}{\sqrt {(1+\lambda)^2+(2-\lambda)^2+(3+\lambda)^2}}\bigg|=\frac{2}{\sqrt 3}$
$\Rightarrow\:\bigg|\large\frac{-2\lambda}{\sqrt {(1+\lambda)^2+(2-\lambda)^2+(3+\lambda)^2}}\bigg|=\frac{2}{\sqrt 3}$
$\Rightarrow\:12\lambda^2=4(1+\lambda^2+2\lambda+4+\lambda^2-4\lambda+9+\lambda^2+6\lambda)$
$\Rightarrow\:16\lambda+56=0$
$\Rightarrow\:\lambda=-\large\frac{7}{2}$
Substituting the value of $\lambda$ in (iii) the required eqn. of the plane is
$5x-11y+z=17$