find the distance between the point and the plane
(2, 5, 3)
3x + 2y - 3z = -6
Short way:
Did you come across this formula yet?
distance from a point (a,b,c) to the plane
Ax + By + Cz + D = 0 is
|Aa + Bb + Cc + D|/√(A^2 + B^2 + C^2)
distance = |6 + 10 - 9 + 6|/√(9+4+9)
= 13/√22
longer way:
parametric equation of perpendicular line from (2,5,3) to the plane is
x = 2 + 3t
y = 5 + 2y
z = 3 - 3t
intersection of this line with the plane:
3(2+3t) + 2(5+2t) - 3(3-3t) = -6
6+9t + 10+4t - 9+9t = -6
22t = -13
t = -13/22
then x = 2-39/22 = 5/22
y = 5-26/22 = 84/22
z = 3+39/22 = 105/22
distance = √( (2-5/22)^2 + (5-84/22)^2 + (3 - 105/22)^2 )
= ...
I will leave it up to you to show this is 13/√22