With [tex]p=12k+1[/tex], it's never the case that [tex]3\equiv0\mod p[/tex], so certainly [tex]\left(\dfrac3p\right)\neq0[/tex].
We can also eliminate the case that [tex]\left(\dfrac3p\right)=-1[/tex] by coming up with a counter-example. Notice that [tex]p=13[/tex] for [tex]k=1[/tex], and that [tex]16=4^2\equiv3\mod13[/tex].
So because there exists some [tex]x[/tex] such that [tex]x^2\equiv3\mod p[/tex] it follows that [tex]\left(\dfrac3p\right)=1[/tex].
