• question_answer Find the equation of the plane passing through the line of intersection of planes $2x+y-z=3,$ $5x-3y+4z+9=0$ and parallel to the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}.$ OR Find the distance of the point $(-\,2,\,\,3,\,\,-4)$ from the line $\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3z+4}{5}$ measured parallel to the plane $4x+12y-3z+1=0.$

 Given planes are$2x+y-z=3$                     ?(i) and       $5x-3y+4z+9=0$                       ?(ii) Also, the given line is $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}$                 ?(iii) Equation of any plane through the intersection of the planes (i) and (ii) is $(2x+y-z-3)+\lambda \,(5x-3y+4z+9)=0$ $\Rightarrow$$(2+5\lambda )\,x+(1-3\lambda )\,y+(4\lambda -1)\,z-(3-9\lambda )=0$ ?(iv) According to the question, plane (iv) is parallel to line (iii). $\therefore$$(2+5\lambda )\cdot 2+(1-3\lambda )\cdot 4+(4\lambda -1)\cdot 5=0$ $\Rightarrow$   $18\lambda +3=0\Rightarrow$$\lambda =\frac{-\,1}{6}$ From Eq. (iv), we have $\left( 2-\frac{5}{6} \right)x+\left( 1+\frac{1}{2} \right)y+\left( \frac{-\,2}{3}-1 \right)z-\left( 3+\frac{3}{2} \right)=0$ $\Rightarrow$               $7x+9y-10z-27=0$ Which is the required equation of plane. OR Given equation of line is $\frac{x+2}{3}=\frac{2\left( y+\frac{3}{2} \right)}{4}=\frac{3\left( z+\frac{4}{3} \right)}{5}$ or         $\frac{x+2}{9}=\frac{y+\frac{3}{2}}{4}=\frac{3\left( z+\frac{4}{3} \right)}{5}$ General point on line is $P\left( 9\lambda -2,\,\,6\lambda -\frac{3}{2},\,\,5\lambda -\frac{4}{3} \right)$          ?(i) DR?s of AP are$9\lambda -2+2,$$6\lambda -\frac{3}{2}-3,$$5\lambda -\frac{4}{3}+4$ i.e         $9\lambda ,$$6\lambda -\frac{9}{2},$$5\lambda +\frac{8}{3}$. If AP is parallel to the plane$4x+12y-3z+1=0$ Then, $36\lambda +12\,\left( 6\lambda -\frac{9}{2} \right)-3\left( 5\lambda +\frac{8}{3} \right)=0$ $\Rightarrow$$36\lambda +72\lambda -54-15\lambda -8=0$ $\Rightarrow$$93\lambda -62=0\Rightarrow \lambda =\frac{2}{3}$ On putting the value of $\lambda$in Eq. (i), we get the point of intersection is $P\,(6-2,\,\,4-\frac{3}{2},\,\,2)$, i.e. $P\,\left( 4,\,\,\frac{5}{2},\,\,2 \right)$ $\therefore$Distance$AP=\sqrt{{{(4+3)}^{2}}+{{\left( \frac{5}{2}-3 \right)}^{2}}+{{(2+4)}^{2}}}$ $=\sqrt{36+\frac{1}{4}+36=}\sqrt{\frac{289}{4}}=\frac{17}{2}units$