Question

The equation of the plane through the line of intersection of $2x-y+3z+1=0,x+y+z+3=0$ and parallel to the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ is:

• A. $x-5y+3z=7$
• B. $x-5y+3z=17$
• C. $x+5y+3z=17$
• D. $x+5y+3z=-7$

Answer 1

The correct option is A $x-5y+3z=7$

Given: $S_1:2x-y+3z+1=0$ and
$S_2:x+y+z+3=0$
Any plane through the line of intersection is $(2x-y+3z+1)+\lambda (x+y+z+3)=0$, using $S_1+\lambda S_2=0$
$(2+\lambda )x+(-1+\lambda )y+(3+\lambda )z+1+3\lambda =0\cdots \left(i\right)$
If it is plane is parallel to the given line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$, then the normal to the plane has to be perpendicular to the above line.
$\therefore (2+\lambda ).1+(\lambda -1).2+(3+\lambda ).3=0(\because l_1{l}_2+m_1{m}_2+n_1{n}_2=0)$
$\therefore \lambda =-\frac{3}{2}$ and
The required plane is $x-5y+3z-7=0$