Note that the determinant of a lower (or upper) triangular matrix is the product of its diagonal elements. Using this fact, we want to create a triangular matrix out of your matrix
\begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix}
So, I will start with the last row and subtract it from the second row to get
\begin{bmatrix} 2 & 3 & 10 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix}
Now, I want to get rid of the $2$ in the first row. I thus multiply the last row by $2$ and subtract it from the first row to obtain:
\begin{bmatrix} 0 & 1 & 16 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix}
Finally, I subtract the second row from the first one to obtain
\begin{bmatrix} 0 & 0 & 15 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix}
We now have
$$\det \begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix} = \det \begin{bmatrix} 0 & 0 & 15 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix}$$
Now, I will transform the RHS matrix to an upper diagonal matrix. I can exchange the first and the last rows. Exchanging any two rows changes the sign of the determinant, and therefore
$$\det \begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix} = -\det \begin{bmatrix} 1 & 1 & -3 \\ 0 & 1 & 1 \\ 0 & 0 & 15 \end{bmatrix}$$
The matrix on the RHS is now an upper triangular matrix and its determinant is the product of its diagonal elements, which is $15$. With the minus sign, the $\det$ of our initial matrix is thus $-15$.
The key idea in using row operations to evaluate the determinant of a matrix is the fact that a triangular matrix (one with all zeros below the main diagonal) has a determinant equal to the product of the numbers on the main diagonal. Therefore one would like to use row operations to 'reduce' the matrix to triangular form.
However, the effect of using the three row operations on a determinant are a bit different than when they are used to reduce a system of linear equations.
(1) Swapping two rows changes the sign of the determinant
(2) When dividing a row by a constant, the constant becomes a factor written in front of the determinant.
(3) Adding a multiple of one row to another does not change the value of the determinant.
Let's apply these operations to your matrix to find its determinant.
First we want to produce two zeros in rows $2$ and $3$ of column $1$. (Remember our goal is to produce all zeros below the main diagonal, and we do this one column at a time beginning with column $1$.)
The two row operations $-R_1+R_2\to R_2$ and $-4R_1+R_3\to R_3$ will accomplish this goal, and will not change the value of the determinant.
\begin{eqnarray}
\begin{vmatrix}1 & 7 & -3\\ 1 & 3 & 1 &\\ 4 & 8 & 1 \end{vmatrix} &=&
\begin{vmatrix} 1 & 7 & -3\\0 & -4 & 4\\0 & -20 & 13\end{vmatrix}
\end{eqnarray}
Now all that remains is to obtain a $0$ in row $3$ column $2$. We see that adding $-5$ times row $2$ to row $3$ will accomplish this. That is, $-5R_2+R_3\to R_3$.
\begin{eqnarray}
\begin{vmatrix}1 & 7 & -3\\ 1 & 3 & 1 &\\ 4 & 8 & 1 \end{vmatrix} &=&
\begin{vmatrix} 1 & 7 & -3\\0 & -4 & 4\\0 & -20 & 13\end{vmatrix}\\
&=& \begin{vmatrix}
1 & 7 & -3\\0 & -4 & 4\\0 & 0 & -7
\end{vmatrix}
\end{eqnarray}
Since we only had to use the third row operation, the one which does not change the value of the determinant and since we now have a triangular matrix, we find the determinant by multiplying the numbers on the main diagonal.
\begin{eqnarray}
\begin{vmatrix}1 & 7 & -3\\ 1 & 3 & 1 &\\ 4 & 8 & 1 \end{vmatrix} &=&
\begin{vmatrix} 1 & 7 & -3\\0 & -4 & 4\\0 & -20 & 13\end{vmatrix}\\
&=& \begin{vmatrix}
1 & 7 & -3\\0 & -4 & 4\\0 & 0 & -7
\end{vmatrix}\\
&=&28
\end{eqnarray}
Best Answer
$$\begin{pmatrix} -1 & 2& -3& 4\\ 5 & 0& 2& -2\\ 2& 1& 1& 2\\ 0& 0& 3& -2 \end{pmatrix}\stackrel{R_2+5R_1\;,\;\;R_3+2R_1}\longrightarrow\begin{pmatrix} -1 & 2& -3& 4\\ 0 & 10& -13& 18\\ 0& 5& -5& 10\\ 0& 0& 3& -2 \end{pmatrix}\stackrel{R_3-\frac12R_2}\longrightarrow$$
$$\begin{pmatrix} -1 & 2& -3& 4\\ 0 & 10& -13& 18\\ 0& 0& 1.5& 1\\ 0& 0& 3& -2 \end{pmatrix}\stackrel{R_4-2R_3}\longrightarrow\begin{pmatrix} -1 & 2& -3& 4\\ 0 & 10& -13& 18\\ 0& 0& 1.5& 1\\ 0& 0& 0& -4 \end{pmatrix}$$
The matrix is now an upper triangular one and its determinant is just the product of its main diagonal's entries...which is certainly $\;60\;$ .