Your row reductions seem fine, thus far. You can row reduce some more:
try to get an upper triangular matrix; short of that, it will simplify the calculation of the determinant.
$(a)$ You can factor out $4$ from the third row.
$$\text{ det}
\begin{pmatrix}
3 & 2 & -1& 4 \\
2 & k & 6 & 5 \\
0 & 4 & 0 & 4 \\
0 & 0 & 4 & -5\\
\end{pmatrix}
=33\iff 4 \text{ det} \begin{pmatrix}
3 & 2 & -1& 4 \\
2 & k & 6 & 5 \\
0 & 1 & 0 & 1 \\
0 & 0 & 4 & -5\\
\end{pmatrix}
=33$$
$$
\text{Add}\;-2(R_3) \text{ to}\;R_1 \implies 4 \text{ det} \begin{pmatrix}
3 & 0 & -1& 2 \\
2 & k & 6 & 5 \\
0 & 1 & 0 & 1 \\
0 & 0 & 4 & -5\\
\end{pmatrix}
= 33$$
$$
\text{Add }\;R_4 \text{ to}\; R_2\implies 4\text{ det} \begin{pmatrix}
3&0&-1&2\\
2&k&10&0\\
0&1&0&1\\
0&0&4&-5\\
\end{pmatrix}
= 33$$
At this point, I'd suggest simply expanding along the column containing $k$; with the additional row reduction, that may simplify the process. Don't worry if you end up with with an equation in which $k$ evaluates to a fraction (decimal)!
$$
\text {det} \begin{pmatrix}
3 & 0 & -1& 2 \\
2 & k & 10 & 0 \\
0 & 1 & 0 & 1 \\
0 & 0 & 4 & -5\\
\end{pmatrix}
= \frac{33}{4}\tag{1}$$
$$k\text{ det}\; \begin{pmatrix}
3 &-1&2\\
0&0&1\\
0&4&-5\\
\end{pmatrix}
- \text{ det}\; \begin{pmatrix}
3&-1&2\\
2&10&0\\
0&4&-5\\
\end{pmatrix}\tag{2}$$
$$= k\text{ det}\; \begin{pmatrix}
3 &-1&2\\
0&0&1\\
0&4&-5\\
\end{pmatrix}
- \left(3\text{ det}\; \begin{pmatrix}
10&0\\
4&-5\\
\end{pmatrix}
-2\text{ det}\; \begin{pmatrix}
-1&2\\
4&-5\\
\end{pmatrix}\right)= \frac{33}{4}
\tag{3}$$
$$-12k - [3(-50) - 2(-3)] = \frac{33}{4}$$
$$-12k +144 = \frac{33}{4}$$
$$k=\frac{181}{16}$$
Note:
In $(1)$, I simply divided both sides of the equation by $4$.
In $(2)$, I expanded along the second column, the column containing $k$. Note the sign of each resulting determinants.
In $(3)$, I expanded along the first column of the second matrix in $2$; again, we need to keep.
The rest is simplification.
$$\mathbf P=\begin{pmatrix}a&2d&1\\b&2e&-2\\c&2f&-1\end{pmatrix},\,\mathbf U=\begin{pmatrix}a&b&c\\2&3&2\\d&e&f\end{pmatrix},\,\mathbf V=\begin{pmatrix}a&b&c\\d&e&f\\1&5&3\end{pmatrix}$$
For later use, we take the transpose of $\mathbf P$:
$$\mathbf P^\top=\begin{pmatrix}a&b&c\\2d&2e&2f\\1&-2&-1\end{pmatrix}$$
Recall that $\det\mathbf P=\det(\mathbf P^\top)$.
Denote by $\mathbf P_{i,j}$ the permutation matrix that, upon multiplication by a matrix $\mathbf A_{m\times n}$, swaps rows $i$ and $j$ in $\mathbf A$. So we can write
$$\mathbf P_{2,3}\mathbf U=\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}\mathbf U=\begin{pmatrix}a&b&c\\d&e&f\\2&3&2\end{pmatrix}$$
Recall that the determinant of any square matrix with any pair of its rows/columns swapped negates the value of the new matrix, so that $\det\mathbf U=-\det(\mathbf P_{i,j}\mathbf U)$ when $i\neq j$.
Now, observe that we can write the last row of $\mathbf V$ as a linear combination of the last rows of $\mathbf P^\top$ and $\mathbf P_{2,3}\mathbf U$:
$$\begin{pmatrix}1&5&3\end{pmatrix}=(-1)\begin{pmatrix}1&-2&-1\end{pmatrix}+\begin{pmatrix}2&3&2\end{pmatrix}$$
The determinant has the property that it is linear with respect to any given row, which is to say: focusing our attention on a single row, if we can write it as a linear combination of other row vectors, then we can expand the determinant as the sum of two component determinants. To illustrate in practice, we can write
$$\begin{vmatrix}a&b&c\\d&e&f\\1&5&3\end{vmatrix}=\begin{vmatrix}a&b&c\\d&e&f\\(-1)1&(-1)(-2)&(-1)(-1)\end{vmatrix}+\begin{vmatrix}a&b&c\\d&e&f\\2&3&2\end{vmatrix}$$
Aside: I highly recommend watching this lecture from MIT if you ever feel the need to brush up on the properties of the determinant. Strang does a great job of explaining them.
Next, we can pull out a factor of $-1$ from the first determinant, and simultaneously distribute a factor of $2$ along the second row of the first matrix,
$$\begin{vmatrix}a&b&c\\d&e&f\\1&5&3\end{vmatrix}=-\frac12\begin{vmatrix}a&b&c\\2d&2e&2f\\1&-2&-1\end{vmatrix}+\begin{vmatrix}a&b&c\\d&e&f\\2&3&2\end{vmatrix}$$
and we see that we have written $\det\mathbf V$ in terms of known determinants. We get
$$\det\mathbf V=-\frac12\det(\mathbf P^\top)+\det(\mathbf P_{2,3}\mathbf U)=-\frac12\det\mathbf P-\det\mathbf U=-5-(-3)=-2$$
Best Answer
The determinant is linear in every row and in every column. Thus, this problem is equivalent to finding $φ(-4,-7,-10)$ for a linear map $φ \colon ℚ^3 → ℚ$ with $φ(1,1,1) = -3$ and $φ(1,2,3)= 5$. Do you see that – what is $φ$ here? Can you solve this reduced problem?