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I am just posting my comment as an answer. For a regular ring $R$ and an ideal $I$ generated by $n$ elements, the embedded primes of $I$ can, indeed, have height strictly larger than $n$. For instance, let $R$ be $k[x,y,u,v,w]$. Let $I$ be the ideal generated by $4$ elements, $$ I = \langle xu^3,yv^3,x^2u^2w-yuvw^2,y^2v^2w-xuvw^2 \rangle. $$ Denote by $S$ the $k$-subalgebra $k[x,y,u,v]$ of $R$. The $S$-submodule of $R/I$ generated by $\overline{1}$ and $\overline{w}$ is $$ \left(S/\langle xu^3,yv^3\rangle\cdot 1 \right) \oplus \left( S/\langle xu^3,yv^3,x^2u^2v^2,y^2u^2v^2\rangle\cdot \overline{w}\right). $$ Thus, the image of $xyu^2v^2w$ in $R/I$ is nonzero. Yet the annihilator equals all of $\langle x,y,u,v,w\rangle$. Thus, the maximal ideal $\langle x,y,u,v,w\rangle$ is an embedded prime of $I$. This maximal ideal has height $5$, which is strictly greater than $4$.