The distance of the point (1,−2,3) from the plane x−y+z−5=0, parallel to the line x2=y3=z−1−6 is

Equation of line through A(1,−2,3) and parallel to the line x2=y3=z−1−6 is x−12=y+23=z−3−6=r………(1) Any point P on this line is P(2r+1,3r−2,−6r+3) .  If P is the point of intersection of line (1) and plane x−y+z−5=0, then 2r+1−3r+2−6r+3−5=0⇒r=17∴P97,−117,157 Required distance =AP=1−972+−2+1172+3−1572=449+949+3649=1