Question

What is the equation of the plane containing the triangle shown?

Answer 1

a 2x + 3y + z = 6

b 3x + 2y + 6z = 6

c 2x + 6y + 3z = 6

d 6x + 3y + 2z = 6

Answer 2

write the plane in the intercept form of x/a +y/b+z/c=1 just write the intercepts and you will get your answers

Answer 3

$$
P=<1,0,0>\\
Q=<0,2,0>\\
R=<0,0,3>
$$
$$
\vec{PQ}=<-1,2,0>\\
\vec{PR}=<-1,0,3>\\
\vec{PQ}\times\vec{PR}=<6,3,2>
$$
Where the last equation is the cross product - http://www.wolframalpha.com/input/?i=%28-1%2C2%2C0%29+cross+%28-1%2C0%2C3%29
This is perpendicular to the plane determined by PQ and PR.

So the equation of the plane is
$$
<6,3,2>\cdot<x-1,y,z>=0\\
=6(x-1)+3y+2z=0\\
6x-6+3y+2z=0\\
6x+3y+2z=6
$$
The first equation follows since <6,3,2> is perpendicular to the plane, which is also perpendicular to the vector <x,y,z> - \(<1,0,0>\), which is a vector on the plane.

See http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx for the theory.