Given four points,4(2, 1,0), B(1,0, 1),C(3,0, 1)and D(0,0,2).Point D lies on a line L orthogonal to the plane determined by the points A, B and C.The equation of the Plane ABC isThe equation of the line I isThe perpendicular distance of D from the plane ABC is

x−2y−1z1−20−11−03−20−11−0=0(x−2)[(−1)−(−1)]−(y−1)[(−1)−1]+z1+1=0or  2(y−1)+2z=0or  y+z−1=0The vector normal to the plane is r→=0i^+j^+k^ The equation of the line through (0, 0, 2) and parallel  to n→ is r→=2k^+λ(j^+k^)The perpendicular distance of D(0, 0, 2) from plane  ABC is 2−112+12=12.x−2y−1z1−20−11−03−20−11−0=0(x−2)[(−1)−(−1)]−(y−1)[(−1)−1]+z1+1=0or  2(y−1)+2z=0or  y+z−1=0The vector normal to the plane is r→=0i^+j^+k^ The equation of the line through (0, 0, 2) and parallel  to n→ is r→=2k^+λ(j^+k^)The perpendicular distance of D(0, 0, 2) from plane  ABC is 2−112+12=12.x−2y−1z1−20−11−03−20−11−0=0(x−2)[(−1)−(−1)]−(y−1)[(−1)−1]+z1+1=0or  2(y−1)+2z=0or  y+z−1=0The vector normal to the plane is r→=0i^+j^+k^ The equation of the line through (0, 0, 2) and parallel  to n→ is r→=2k^+λ(j^+k^)The perpendicular distance of D(0, 0, 2) from plane  ABC is 2−112+12=12.