SOLUTION: Find the equation of the plain passing through the points P(2, -3, 1), P’(5, -3, -5) and perpendicular to the plane x - 2y + 5z + 20 = 0.

Normal vectors to the two planes must have dot product 0 in order
for them to be perpendicular.

The coefficients of x, y and z in the equation of a plane 
are the components of a vector normal to that plane.

Let the desired plane have an equation



then it will have a normal vector < a, b, c >

The two given points must satisfy the equation of the
desired plane.  So we have these two equations:

a(2)+b(-3)+c(1)+d=0
a(5)+b(-3)+c(-5)+d=0

which simplify to

2a-3b+ c+d=0
5a-3b-5c+d=0

The coefficients of x, y and z of the plane  x - 2y + 5z + 20 = 0
are the components of a normal vector to that plane.  And the dot
product of vectors normal to them must be 0.

< 1, -2, 5 > • < a, b, c > = 0

1a - 2b + 5c = 0

So we have the system of 
equations:



Solve that by the Gauss-Jordan method



Use your TI-84 to get the rref of that matrix, which is



Which tells us



or



So one equation for the desired plane could be



We may multiply through by 11



And then divide through by d

   

That's the best answer.

Edwin