-6x +  5y + 6z = -5
 3x +  2y + 6z = -5
-6x + 23y + hz =  k

For this to have infinitely many solutions, the coefficient determinant must
equal 0:



We evaluate the determinant by copying over the first two columns on the
right



Add the products of the three upper left to lower right diagonal elements:

(-6)(2)(h)+(5)(6)(-6)+(6)(3)(23)
        -12h-180+414
          -12h+234

Now subtract the products of the three upper right to lower left diagonal elements:

-(6)(2)(-6)-(-6)(6)(23)-(5)(3)(h)
          72+828-15h
           900-15h

We combine those and get -12h+234+900-15h = -27h+1134

Set that equal to 0

-27h+1134 = 0
     -27h = -1134
        h = 42

Now our system becomes:

-6x +  5y +  6z = -5
 3x +  2y +  6z = -5
-6x + 23y + 42z =  k

Eliminate z from the first two by multiplying the
1st equation by -1 and adding.  That gives:

        9x - 3y = 0

Now eliminate z from the second and third by
multiplying the second by -7 and adding.  

  -21x - 14y - 42z = 35
   -6x + 23y + 42z =  k
---------------------------
  -27x +  9y       = 35+k

Now we multiply the equation 9x - 3y = 0 through
by 3 and add it to that result:

        -27x +  9y = 35+k
         27x -  9y = 0
-----------------------------
                 0 = 35+k
                35 = k

So h = 42 and k = 35

That's the answer.  But eventually, you'll have to give the solution
in terms of z.  When you do, the solution will be:

The solution is (x,y,z) = 

Edwin