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

first, you divide the triangle in two, tô form a retangle triangle. after, using pythagoream theorem:

18^2 = 14^2 + (x/2)^2

(x^2/4) + 196 = 324

x^2/4 = 324 - 196

x^2/4 = 128

x^2 = 512

x = √512

x = √2^9

x = 16√2 cm

i hope i you.