Question 1158864
<pre>
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

{{{ax+by+cz+d=0}}}

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:

{{{system(2a-3b+ c+d=0,
5a-3b-5c+d=0,a - 2b + 5c = 0)}}}

Solve that by the Gauss-Jordan method

{{{(matrix(3,6,

2,-3,1,1,"|",0,
5,-3,-5,1,"|",0,
1,-2,5,0,"|",0))}}}

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

{{{(matrix(3,6,

1,0,0,-4/11,"|",0,
0,1,0,-7/11,"|",0,
0,0,1,-2/11,"|",0))}}}

Which tells us

{{{system(a-expr(4/11)d=0,b-expr(7/11)d=0, c-expr(2/11)d=0)}}}

or

{{{system(a=expr(4/11)*d,b=expr(7/11)*d, c=expr(2/11)*d)}}}

So one equation for the desired plane could be

{{{expr(4/11)*d*x+expr(7/11)*d*y+expr(2/11)*d*z+d=0}}}

We may multiply through by 11

{{{4d*x+7d*y+2d*z+11*d=0}}}

And then divide through by d

{{{4x+7y+2z+11=0}}}   

That's the best answer.

Edwin</pre>