$3.$The Python code for the pseudocode: k $=($j $+13)/27$loop: if k $>10$ then goto out k $=$ k $+1$ i $=3*$ k $-1$ goto loopout: $...$can be written to be: $($a$).$ Do K $=($J $+13)/27,10$ I $=3*($K $+1)-1$ End Do$($b$).$ for k in $($j $+13)/27..10$ loop i :$=3*($k $+1)-1$; end loop;$($c$).$ for k in $($j $+13)/27..10$ loop i :$=3*($k $+1)-1$; end loop;a$.($a$)$ is correctb.$($b$)$ is correct.c$.($c$)$ is correctd.Both $($a$)$ and $($c$)$ are correct.e$.$None of the choice is correct.