JEE Main 2014 — Three Dimensional Geometry Question with Solution

Equation of the plane containing the line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ is $$a(x-1)+b(y-2)+c(z-3)=0$$ where $a.1+b.2+c.3=0$ i.e., $a+2b+3c=0$ (ii) Since the plane (i) parallel to the line $$\begin{array}{ll}& \frac{x}{1}=\frac{y}{1}=\frac{z}{4}\\ \therefore & a.1+b.1+c.4=0\\ \text{ i.e., }& \mathrm{a}+\mathrm{b}+4\mathrm{c}=0\end{array}$$ (iii) From (ii) and (iii), $$\begin{array}{c}\frac{a}{8-3}=\frac{b}{3-4}=\frac{c}{1-2}=k\text{ (let) }\\ \therefore a=5k,b=-k,c=-k\end{array}$$ On putting the value of $a,b$ and $c$ in equation (i), $$\begin{array}{rl}& 5(x-1)-(y-2)-(z-3)=0\\ & \Rightarrow 5x-y-z=0\end{array}$$ when $x=1,y=0$ and $z=5$; then L.H.S. of equation (iv) $=5x-y-2$ $$\begin{array}{rl}& =5\times 1-0-5\\ & =0\\ & =\text{ R.H.S. of }\end{array}$$ equation (iv) Hence coordinates of the point $(1,0,5)$ satisfy the equation plane represented by equations (iv), Therefore the plane passes through the point $(1,0,5)$