Answer
See below
Work Step by Step
Given the set of vectors $\{(1,-3,2),(1,0,-1),(1,2,-4)\}$
Obtain: $\begin{vmatrix}
1 & 1 & 1\\ -3 & 0 & 2 \\ 2 & -1 & -4
\end{vmatrix}=1\begin{vmatrix}
0 & 2 \\
-1 & -4
\end{vmatrix}-11\begin{vmatrix}
-3 & 2 \\
2 & -4
\end{vmatrix}+1.1\begin{vmatrix}
3 & 0\\
2 & -1
\end{vmatrix}=1.2-1.8+1.3=-3 \ne 0$
Thus, the set of vectors are notcoplanar and therefore set $\{(1,-3,2),(1,0,-1),(1,2,-4)\}$ does span $R^3$
Rewrite $v=(9,8,7)=a(1,-3,2)+b(1,0,-1)+c(1,2,-4)$
We have the system:
$a+b+c=9\\
-3a+2c=8\\
2a-b-4c=7$
To solve the system, we will form the matrix $\begin{bmatrix}
1 & 1 & 1 |9 \\
-3 & 0 & 2|8\\
2 & -1 & -4 |7
\end{bmatrix}\approx \begin{bmatrix}
1 & 1 & 1 |9 \\
-3 & 0 & 2|8\\
3 & 0 & -3 |16
\end{bmatrix}\approx\begin{bmatrix}
1 & 1 & 1 |9 \\
-3 & 0 & 2|8\\
0 & 0 & -1 |24
\end{bmatrix}\approx \begin{bmatrix}
1 & 1 & 1 |9 \\
-3 & 0 & 0|56 \\
0 & 0 & -1 |24
\end{bmatrix}\approx\begin{bmatrix}
1 & 1 & 0 | 33 \\
-3 & 0 & 0|56\\
0 & 0 & -1 |24
\end{bmatrix} \approx \begin{bmatrix}
1 & 1 & 0 | 33 \\
1 & 0 & 0|-\frac{56}{3}\\
0 & 0 & -1 |24
\end{bmatrix}\approx \begin{bmatrix}
0 & 1 & 0 | \frac{155}{33}\\
1 & 0 & 0|-\frac{56}{3}\\
0 & 0 & -1 |24
\end{bmatrix}$
Hence, $a=-\frac{56}{3}\\
b=\frac{155}{3}\\
c=-24$
Hence vector $v=(5,-7)$ can be written as a linear combination of vectors $\{-\frac{56}{3}(1,-3,2)+\frac{155}{3}(1,0,-1)-24(1,2,-4)\}$
