= Solution
Use $[\lambda]$ for the ordinary <irreducible> <character> of $S_{|\lambda|}$. Multiplication of different degrees is the product in the <induction ring of symmetric-group characters>: it induces an <external tensor product of group representations>. For $\alpha\subseteq\nu$, the <skew Young diagram> $\nu/\alpha$ is the set of cells of $\nu$ outside $\alpha$. A filling of it is a <semistandard Young tableau> when entries weakly increase along rows and strictly increase down columns. Its content $\beta$ means that $i$ occurs $\beta_i$ times. Its reading word takes each row right to left and the rows top to bottom. A <lattice word> has, in every initial segment, at least as many $i$'s as $(i+1)$'s for every $i$.
The <Littlewood–Richardson rule> says
$$
[\alpha][\beta]=\sum_{\nu\vdash |\alpha|+|\beta|}c_{\alpha\beta}^\nu[\nu],
$$
where the <Littlewood–Richardson coefficient> $c_{\alpha\beta}^\nu$ counts such skew semistandard fillings with content $\beta$ and lattice reading word. If $\alpha$ is not contained in $\nu$, the coefficient is zero. Over <characteristic zero> this is an actual direct-sum multiplicity statement; over other <fields> the analogous statement describes a <Specht filtration>, not necessarily a decomposition into simple <modules>.
For the required coefficient, the five skew cells are
$$
(1,4),(1,5),\quad(2,3),\quad(3,2),\quad(4,1).
$$
Their columns are all distinct. Content $(3,2)$ uses three $1$'s and two $2$'s. The first reading letter must be $1$ by the <lattice word> condition, and row monotonicity forces both top cells to be $1$. The remaining three cells, in reading order, can then be filled by $122$, $212$, or $221$. The resulting words $11122$, $11212$, $11221$ all satisfy the lattice inequalities, and exhaust the possibilities. Thus
$$
\boxed{c_{(3,2,1),(3,2)}^{(5,3,2,1)}=3.}
$$
For the <determinantal form of a symmetric-group character>, set $[0]=1$ in degree zero and $[r]=0$ for $r<0$. The one-row <character> $[r]$ is trivial. The one-row specialization of the <Littlewood–Richardson rule> is the <Pieri rule>: multiplication by $[r]$ adds a <horizontal strip> of $r$ cells, since all entries are $1$ and no column can contain two of them.
Here is an explicit <determinant> argument, so that the desired form is genuinely deduced. Under the <Frobenius characteristic map>, $[r]$ corresponds to the <complete homogeneous symmetric polynomial> $h_r$, whose monomials enumerate weakly increasing sequences of $r$ entries. Fix $N$ variables and $l$ rows. Let a directed lattice path run right or up from $A_i=(-i,1)$ to $B_j=(\lambda_j-j,N)$, and assign weight $x_k$ to a horizontal step at height $k$ and weight one to a vertical step. The sum of its weights is $h_{\lambda_j-j+i}$. Expanding the <determinant> of this path matrix sums signed families connecting the sources to a <permutation> of the destinations.
Cancel intersecting families as follows. Choose the first intersection in a fixed order of vertices, and then the first pair of paths meeting there. Exchange their tails after that vertex. The total weight is unchanged, while the endpoint <permutation> changes by a <transposition>, reversing the sign. Choosing the intersection in this fixed order makes the operation an involution. The surviving families are nonintersecting; the ordered sources and ordered destinations force the identity endpoint pairing. Record the heights of the horizontal steps of path $i$ as row $i$ of a tableau. Rows weakly increase, and nonintersection of successive paths is precisely strict increase down the shared columns. Conversely every <semistandard Young tableau> gives one surviving path family. Thus the <determinant> has the semistandard-tableau generating function $s_\lambda$.
<Young's rule> identifies this generating function with the characteristic of $[\lambda]$: its coefficient at a monomial of weight $\mu$ is $K_{\lambda\mu}$, the multiplicity obtained by pairing with the <Young permutation module> of type $\mu$. The <Frobenius characteristic map> is injective and respects induction products. Transposing the path matrix does not change its <determinant>, so
$$
\boxed{[\lambda]=\det\bigl([\lambda_i+j-i]\bigr)_{1\le i,j\le l}.}
$$
This is the <character> version of the <Jacobi–Trudi identity>.
For the example the matrix is
$$
[3,2,2]=\det\begin{pmatrix}[3]&[4]&[5]\\{}[1]&[2]&[3]\\{}[0]&[1]&[2]\end{pmatrix}.
$$
Expanding gives the concise answer
$$
\boxed{[3,2,2]=[3][2]^2-[3]^2[1]-[4][1][2]+[4][3]+[5][1]^2-[5][2].}
$$
At the identity the six induction-product <dimensions> are $210,140,105,35,42,21$, respectively; their alternating sum is $21$, agreeing with the <hook-length formula> for $(3,2,2)$.
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