For a certain value of k, the system

The third equation has no y-term, so let's eliminate the y-term from the combination of the first two equations.

x + y + 3z  =  10     --->     x -2     --->     -2x - 2y - 6z  =  -20

-4x + 2y + 5z  =  7       --->                      -4x + 2y + 5z  =  7

Add down the columns:                             -6x - z  =  -13   

Since  kx + z  =  3        --->         z  =  3 - kx

Substituting these:        -6x - (3 - kx)  =  -13

                                    -6x - 3 + kx  =  -13

                                      6x + 3 - kx  =  13

                                          6x - kx  =  10

                                          x(6 - k)  =  10

If  6 - k  =  0,  there is no value for x that will result in a product of 10,

so  k = 6  results in no possible solution.

For every other value of k, there will be a value of x that can produce a product of 10.

Here's another way to tackle this one - although I still probably like geno's method better !!!!

Get y by itself in the first equation

x + y + 3z = 10      →   y =  10 - x - 3z

Substitute this into the second equation and simplify:

-4x + 2y +5z = 7    →   -4x + 2(10 - x - 3z) + 5z = 7   →   -4x + 20 -2x - 6z + 5z  = 7  →  -6x - z = -13  →

6x + z  = 13

So we have this system:

k*x + z = 3

6x + z = 13 

And notice that, if k =6, we would have this system:

6x + z  = 3      and

6x + z  = 13       but this is impossible, because two things added together can't possibly have two different outcomes 

So....as geno found, when k = 6, the system has no solution...!!!