pls_help!!!Find the equation of the plane through (2,2,1) that cuts off the smallest volume of the 1st octant?

Question

Find the equation of the plane through (2,2,1) that cuts off the smallest volume of the 1st octant..... pls, help...I'll need explanations, not that the answer on this one..thanks!

Answer 1

a lazy answer is like this
x+y+z = 5

edit
an unlazy answer is shown below

eq. of the plane
ax + by + cz = 2a + 2b + c

the plane intersects each axis at
X = ((2a + 2b + c)/a,0,0)
Y = (0,(2a + 2b + c)/b,0)
Z = (0,0,(2a + 2b + c)/c)

V = XYZ/6
= (2a + 2b + c)³/6abc

V minimizes at 2a=2b=c,
so if we let a=1, then b=1 and c=2

answer
x + y + 2z = 6

Answer 2

The equation of the plane through P(2,2,1) that cuts off the smallest volume of the 1st octant is:

x + y + 2z = 6

Answer 3

This will obviously be a tetrahedron. Because it will be a tetrahedron lying in the first octant, it has points

(2,0,0), (0,2,0), (0,0,1).
P(2,0,0), Q(0,2,0), R(0,0,1)

PQ = <-2, 2, 0>
PR = <-2, 0, -1>

Take cross product.

PQ X PR = -2i - (2)j +(0 - (-4))k
PQ X PR = -2i - 2j + 4k
Your normal vector to the plane = <-2,-2,4>

-2(x-2) -2(y-2) + 4(z-1) = 0
-2x + 4 - 2y + 4 + 4z - 4 = 0
-2x - 2y + 4z = -4
-x - y + 2z = -2
x + y - 2z = 2.