Question 1075059
Typo suspected.
Did you need to know the values of k
that caused the system to have one solution?
As posted, the system has one solution for each value of k.
There is no value of k that causes a different result.
 
If your first equation was really
{{{2x+y+2=0}}} , 
and your last equation was really
{{{x+y+2=0}}} ,
subtracting the last one from the first one,
you get {{{x=0}}} .
Substituting that value for {{{x}}} in either equation,
you get {{{y+2=0}}} ---> {{{y=-2}}} .
Then, substituting in the second equation
the values found for {{{x}}} and {{{y}}} ,
you get {{{-2k+2z=0}}} ---> {{{z=k}}} .