Find an equation of the plane. the plane that passes through the line of intersection of the planes x − z = 3 and y + 2z = 3 and is perpendicular to the plane x + y − 3z = 6

A plane passing through the line of intersection of planes (x-z-3=0) and (y+2z-3=0) will have an equation that is a linear combination of these:

... (x-z-3) +k(y+2z-3) = 0

The direction vector of this plane is found from the coefficients of (x, y, z) as

... (1, k, -1+2k)

Since we want the plane that is perpendicular to the given one with direction vector (1, 1, -3), so the dot product of these direction vectors will be zero.

... (1, k, -1+2k)•(1, 1, -3) = 0

... 1·1 + k·1 + (-1+2k)·(-3) = 0

... 1 + k + 3 -6k = 0

... 4 = 5k

... k = 4/5

Substituting this into the plane's equation above, we have

... (x-z-3) +(4/5)(y+2z-3) = 0

... 5x -5z -15 +4y +8z -12 = 0 . . . . . . multiply by 5 to eliminate fractions

... 5x +4y +3z = 27 . . . . . . . . . . . . . . the equation of the desired plane

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!