If the line x−23=y−1−5=z+22 lies in the plane x+3y−αz+β=0 then α,β=

The condition that the line  x−x1l=y−y1m=z−z1nto lies on the plane  ax+by+cz+d=0 is al+bm+cn=0and ax1+by1+cz1+d=0Hence,             13+3−5−α2=03−15−2α=02α=−12α=−6And 12+31+6−2+β=05−12+β=0β=7            Therefore, α,β=−6,7