JEE Main & Advanced
JEE Main Paper Phase-I (Held on 09-1-2020 Evening)
question_answer
If the distance 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 \[\frac{x+3}{2}=\frac{y+2}{6}=\frac{z-1}{\lambda }(\lambda \in R)\] is equal to \[\frac{k}{\sqrt{633}},\] then k is equal to [JEE MAIN Held on 09-01-2020 Evening]