As given, the vector from the origin to the point (2,−1,3) is normal to plane. ∴‌‌n=<2,−1,3> Then, general equation of plane 2x−y+3z=d ∵‌‌(2,−1,3) lies on the plane. ∴‌‌2(2)−(−1)+3×3=d⇒d=14 Hence, equation of the plane 2x−y+3z−14=0
