JEE Main 2021 (Online) 27th July Evening ShiftQuestion 1
For real numbers $\alpha$ and $\beta\ne0$, if the point of intersection of the straight lines $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}$ and $\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}$, lies on the plane $x+2y-z=8$, then $\alpha-\beta$ is equal to :
3D Geometry
Solution
Line 1: $(\alpha+t,\,1+2t,\,1+3t)$; line 2: $(4+\beta s,\,6+3s,\,7+3s)$. $y,z$: $1+2t=6+3s$, $1+3t=7+3s\Rightarrow t=1,\ s=-1$. Point: $(\alpha+1,3,4)$. On plane: $\alpha+1+6-4=8\Rightarrow\alpha=5$. $x$: $\alpha+1=4-\beta\Rightarrow\beta=-2$. $\alpha-\beta=7$. Option (C).