JEE Main 2020 (Online) 9th January Evening ShiftQuestion 1
If the distance between the plane, $23x-10y-2z+48=0$ and the plane containing the lines $\frac{x+1}2=\frac{y-3}4=\frac{z+1}3$ and $\frac{x+3}2=\frac{y+2}6=\frac{z-1}\lambda$ ($\lambda\in\mathbf{R}$) is equal to $\frac k{\sqrt{633}}$, then $k$ is equal to ________.
3D Geometry
Answer$3$
Solution
Lines through $(-1,3,-1)$ with direction $(2,4,3)$ and through $(-3,-2,1)$ with direction $(2,6,\lambda)$ are coplanar: $\begin{vmatrix}-2&-5&2\\2&4&3\\2&6&\lambda\end{vmatrix}=0\Rightarrow\lambda=-7$. Normal $=(2,4,3)\times(2,6,-7)=(-46,20,4)\parallel(23,-10,-2)$. Plane: $23x-10y-2z+d=0$ through $(-1,3,-1)$: $-23-30+2+d=0\Rightarrow d=51$. Distance to $23x-10y-2z+48=0$: $\dfrac{|51-48|}{\sqrt{529+100+4}}=\dfrac{3}{\sqrt{633}}$, so $k=3$.