Products of 3- and 4-vectors: consistency of the metric signature

EDIT for clarification of question: Is it correct that the square of a 3-vector picks up a minus sign in the signature ${+—}$? If not, doesn’t that mean the ${-+++}$ convention is preferred (better) because it gives the right answer for products of 3- and 4-vectors?

Let the metric be the Minkowski metric in signature ${+—}$ with $c=1$. For some 4-vector $v^mu=(v^0,v^1,v^2,v^3)$, we have

begin{align}
v^2&=v^mu v^nueta_{munu}
&=v^mu big( v^0eta_{mu0} + v^1eta_{mu1} + v^2eta_{mu2} + v^3eta_{mu3} big)
&=v^mu big( v^0delta_{mu0} -v^1delta_{mu1}- v^2delta_{mu2} -v^3delta_{mu3} big)
&=v^mu big( v^0 -v^1- v^2 -v^3 big)~~.
end{align}

Noting that the individual components of $v$ are scalars multiplied by the four unit vectors $hat e^mu$, we define $v_mu=(v^0,-v^1,-v^2,-v^3)$ so that

begin{align}
v^2&=v^mu v_mu
&=v^0v_0+v^1v_1+v^2v_2+v^3v_3
&=big(v^0big)^2-big(v^1big)^2-big(v^2big)^2-big(v^3big)^2~~.
end{align}

However, if we consider a 3-vector $p^k=(v^1,v^2,v^3)$ such that $v^mu=v^0+p^k$, we have

begin{align}
p^2&=p^k p^jeta_{kj}
&=p^kbig( v^1eta_{k1} + v^2eta_{k2} + v^3eta_{k3} big)
&=p^k big( -v^1delta_{k1}- v^2delta_{k2} -v^3delta_{k3} big)
&=p^k big( -v^1- v^2 -v^3 big)~~.
end{align}

for which we would define $p_k=(-v^1,-v^2,-v^3)$ so that

begin{align}
p^2&=p^k p_k
&=v^1v_1+v^2v_2+v^3v_3
&= -big(v^1big)^2-big(v^2big)^2-big(v^3big)^2~~.
end{align}

However, this is not the correct answer. The correct answer is

$$p^2=vec pcdot vec p= big(v^1big)^2+big(v^2big)^2+big(v^3big)^2~~,$$

where we have implicitly taken the $eta_{jk}$ part of $eta_{munu}$ to have signature ${+++}$. What is going on here? How is this reconciled?