Answer
$\det A=84$
Work Step by Step
Given $$\begin{array}{c}{A=\left[\begin{array}{ccc}
7 & {-1} & 3&4 \\
14 & {2} & 4&6\\
21&1 &3& 4\\
-7&4 &5 &8\end{array}\right] }\end{array} $$
So, we get
$$A=\begin{vmatrix}
7 & {-1} & 3&4 \\
14 & {2} & 4&6\\
21&1 &3& 4\\
-7&4 &5 &8
\end{vmatrix} \xrightarrow{R_2-2R_1,R_3-3R_{1},R_4+R_1} \begin{vmatrix}
7 & {-1} & 3&4 \\
0 & {4} & -2&-2\\
0&4 &-6& -8\\
0&3 &8 &12
\end{vmatrix} \xrightarrow{R_{3}-R_{2},R_4-\frac{3}{2}R_2,\frac{9}{8}R_4+\frac{19}{8}R_3 }\begin{vmatrix}
7 & {-1} & 3&4 \\
0 & {4} & -2&-2\\
0&0 &-4& -6\\
0&0 &0 &-\frac{3}{4}
\end{vmatrix} $$
\begin{array}{l}{\text { Determinant of triangular matrix is the product of its diagonal elements, so: }} \\ {\text {det }A=7 \cdot 4\cdot(-4)\cdot (-\frac{3}{4})=84}\end{array}