The number of solutions of the system of equations

$(d) 0$

The given system of equations can be written in matrix form as follows:]

$\left[\begin{array}{ccc}2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & -3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}7 \\ 1 \\ 5\end{array}\right]$

$A X=B$

Here,

$A=\left[\begin{array}{ccc}2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & -3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}7 \\ 1 \\ 5\end{array}\right]$

Now,

$|A|=2(9-8)-1(-3-2)-1(4+3)$

$=2+5-7$

$=0$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}-3 & 2 \\ 4 & -3\end{array}\right|=1, C_{12}=(-1)^{1+2}\left|\begin{array}{cc}1 & 2 \\ 1 & -3\end{array}\right|=5, C_{13}=(-1)^{1+3}\left|\begin{array}{cc}1 & -3 \\ 1 & 4\end{array}\right|=7$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}1 & -1 \\ 4 & -3\end{array}\right|=-1, C_{22}=(-1)^{2+2}\left|\begin{array}{cc}2 & -1 \\ 1 & -3\end{array}\right|=-5, C_{23}=(-1)^{2+3}\left|\begin{array}{cc}2 & 1 \\ 1 & 4\end{array}\right|=-7$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}1 & -1 \\ -3 & 2\end{array}\right|=-1, C_{32}=(-1)^{3+2}\left|\begin{array}{cc}2 & -1 \\ 1 & 2\end{array}\right|=-5, C_{33}=(-1)^{3+3}\left|\begin{array}{cc}2 & 1 \\ 1 & -3\end{array}\right|=-7$

adj $A=\left[\begin{array}{ccc}1 & 5 & 7 \\ -1 & -5 & -7 \\ -1 & -5 & -7\end{array}\right]^{T}$

$=\left[\begin{array}{lll}1 & -1 & -1 \\ 5 & -5 & -5 \\ 7 & -7 & -7\end{array}\right]$

$\Rightarrow(\operatorname{adj} A) B=\left[\begin{array}{lll}1 & -1 & -1 \\ 5 & -5 & -5 \\ 7 & -7 & -7\end{array}\right]\left[\begin{array}{l}7 \\ 1 \\ 5\end{array}\right]$

$=\left[\begin{array}{c}7-1-5 \\ 35-5-25 \\ 49-7-35\end{array}\right]$

$=\left[\begin{array}{l}1 \\ 5 \\ 7\end{array}\right] \neq 0$

So, the given system of equations has no solution.