The distance of the point (1, -2, 3) from the plane x−y+z=5 measured parallel to the line x2=y3=z−1−6 is

[a] Equation of the line through (1, -2, 3) parallel to the line x2=y3=z−1−6 is x−12=y+23=z−1−6=r(say)       ?..(1) Then any point on (1) is (2r+1,3r−2,−6r+3)if this point lies on the plane x−y+z=5 then (2r+1)−(3r−2)+(−6r+3)=5⇒r=17 Hence the point is (97,−117,157) Distance between (1,−2,3) and (97,−117,157) =(449+949+3649)−−−−−−−−−−−−−−−√=(4959)−−−−−√=1