$\newcommand\tr{\operatorname{tr}}$Let 
$$Q(B):=(Y-XB)^T(Y-XB).$$
Since the column spaces of the matrices $X^TX$ and $X^T$ are the same, there is a matrix $B_*$ such that 
$$X^TXB_*=X^T Y.$$
For each $z\in\Bbb R^{p\times1}$, by (say) differentiating the convex function $B\mapsto z^T Q(B)z$ in $B$, we see that $B_*$ is a minimizer of this function. So, $B_*$ is a minimizer of $Q(B)$ in $B$ wrt to the [Loewner ordering][1]. So, $B_*$ is a minimizer of both $\tr Q(B)$ and $\det Q(B)$ in $B$. 

Generically, $\tr Q(B)$ is strictly convex in $B$, and then $B_*$ is a unique minimizer of $\tr Q(B)$. However, a minimizer of $\det Q(B)$ is not unique.

For instance, suppose that 
$$X=\left(
\begin{array}{cc}
 1 & 3 \\
 1 & 1 \\
\end{array}
\right), \quad
Y=\left(
\begin{array}{cc}
 3 & 0 \\
 2 & 1 \\
\end{array}
\right).$$
Then 
$$B_*=\frac12\,\left(
\begin{array}{cc}
 3 & 3 \\
 1 & -1 \\
\end{array}
\right)$$
is the unique minimizer of $\tr Q(B)$. However, 
$$\left(
\begin{array}{cc}
 0 & 0 \\
 0 & -1 \\
\end{array}
\right)$$
is a minimizer of $\det Q(B)$.



  [1]: https://en.wikipedia.org/wiki/Loewner_order