Given equations are:

$ax+2y–5=0$

$3x+y–1=0$

Above equations are of the form

${{a}_{1}}x+{{b}_{1}}y-{{c}_{1}}=0$

${{a}_{2}}x+{{b}_{2}}y-{{c}_{2}}=0$

Now, ${{a}_{1}}=k,{{b}_{1}}=2,{{c}_{1}}=-5$

${{a}_{2}}=3,{{b}_{2}}=1,{{c}_{2}}=-1$

According to the question,

For the unique solution, the condition is

${{a}_{1}}/{{a}_{2}}\ne {{b}_{1}}/{{b}_{2}}$

$a/3\ne 2/1$

$\Rightarrow a\ne 6$

Therefore, given equations will have unique solution for all real values of a other than $6$.