Question

find the distance of the point (2,1,-1) from the plane 6x-3y+2z-14=0​

Answer 1

Answer:

Find the distance of a point (2, 3, − 5) from the plane x + 2 y − 2 z = 9. View solution A plane passes through ( 1 , − 2 , 1 ) and is perpendicular to two planes 2 x − 2 y + z = 0 and x − y + 2 z = 4

Step-by-step explanation:

Answer 2

Answer:

Distance = | 6(0) - 3(0) + 2(0) - 14 | / √36+9+4 = 14/7