Note: This refers to the system as original given.
Rewrite the equations:
$$\begin{array}{rcccccl}
&&3y&-&4z&=&0\\
2x&-&4y&+&3z&=&0\\
2x&+&y&-&3z&=&0 \\
x&+&y&+&z&=&1
\end{array}$$
Now simply solve the system:
$$\begin{align*}
\left(\begin{array}{rrr|c}
0 & 3 & -4 & 0\\
2 & -4 & 3 & 0\\
2 & 1 & -3 & 0\\
1 & 1 & 1 & 1
\end{array}\right) &\to \left(\begin{array}{rrr|r}
1 & 1& 1 & 1\\
0 & 3 & -4 & 0\\
2 & -4 & 3 & 0\\
2 & 1 & -3 & 0
\end{array}\right) \to\left(\begin{array}{rrr|r}
1 & 1 & 1 & 1\\
0 & 3 & -4 & 0\\
0 & -6 & 1 & -2\\
0 & -1 & -5 & -2
\end{array}\right)\\
&\to \left(\begin{array}{rrr|r}
1 & 1 & 1 & 1\\
0 & 1 & 5 & 2\\
0 & 3 & -4 & 0\\
0 & -6 & 1 & -2
\end{array}\right) \to \left(\begin{array}{rrr|r}
1 & 1 & 1 & 1\\
0 & 1& 5 & 2\\
0 & 0 & -19 & -6\\
0 & 0 & 31 & 10
\end{array}\right).
\end{align*}$$
The last two lines tell us that the system is inconsistent, so the system has no solution.
Alternatively: plugging in $\frac{4}{3}z$ for $y$, we obtain from the second equation
$$\begin{align*}
2x - \frac{16}{3}z + 3z &= 0\\
2x -\frac{7}{3}z &= 0\\
6x -7z &=0.\end{align*}$$
The third equation gives
$$\begin{align*}
2x +\frac{4}{3}z - 3z &= 0\\
2x -\frac{5}{3}z &=0\\
6x - 5z&=0
\end{align*}$$
Since $6x=7z$ and $6x=5z$, then $x=z=0$, hence $y=0$, which makes $x+y+z=1$ impossible.
If the system is meant to be
$$\begin{array}{rcccccl}
-4x&+&3y&&&=&0\\
2x&-&4y&+&3z&=&0\\
2x&+&y&-&3z&=&0 \\
x&+&y&+&z&=&1
\end{array}$$
then proceed as above to solve the system.
$$\begin{align*}
\left(\begin{array}{rrr|r}
-4 & 3 & 0 & 0\\
2 & -4 & 3 & 0\\
2 & 1 & -3 & 0\\
1 & 1 & 1 & 1
\end{array}\right) &\to \left(\begin{array}{rrr|r}
1 & 1 &1 & 1\\
-4 & 3 & 0 & 0\\
2 & -4 & 3 & 0\\
2 & 1 & -3 & 0
\end{array}\right) \to \left(\begin{array}{rrr|r}
1 & 1 & 1 & 1\\
0 & -5 & 6 & 0\\
2 & -4 & 3 & 0\\
0 & 5 & -6 & 0
\end{array}\right)\\
&\to \left(\begin{array}{rrr|r}
1 & 1 & 1 & 1\\
0 & -5 & 6 & 0\\
0 & -6 & 1 & -2
\end{array}\right) \to \left(\begin{array}{rrr|r}
1 & 1 & 1 & 1\\
0 & 1 & 5 & 2\\
0 & -5 & 6 & 0
\end{array}\right)\\
&\to \left(\begin{array}{rrr|r}
1 & 1 & 1 & 1\\
0 & 1 & 5 & 2\\
0 & 0 & 31 & 10
\end{array}\right) \to
\left(\begin{array}{rrr|r}
1 & 0 & -4 & -1\\
0 & 1 & 5 & 2\\
0 & 0 & 1 & \frac{10}{31}
\end{array}\right)\\
&= \left(\begin{array}{rrr|r}
1 & 0 & 0 & \frac{9}{31}\\
0 & 1 & 0 & \frac{12}{31}\\
0 & 0 & 1 & \frac{10}{31}
\end{array}\right).
\end{align*}$$
So the solution is $x = \frac{9}{31}$, $y=\frac{12}{31}$, $z=\frac{10}{31}$.