Question 946224
Let P=(3,0,0), Q=(0,6,0), and R=(0,0,6).
Find PQ and PR, two vectors that lie in the plane.
PQ=(-3,6,0)
PR=(-3,0,6)
Now find the cross product to have a vector normal to the plane.
PQ X PR= (36,18,18)=(2,1,1)
So the plane has the form,
{{{2x+y+z=D}}}
Now use any point to solve for D.
{{{2(3)+0+0=D}}}
{{{D=6}}]
.
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{{{2x+y+z=6}}}