You can divide the equations, to get
$$\begin{align}
\frac1x + \frac1y &= \frac56 = \frac{35}{42}\\
\frac1y+\frac1z &= \frac{10}{21} = \frac{20}{42}\\
\frac1x+\frac1z &= \frac{9}{14} = \frac{27}{42}
\end{align}$$
Subtracting the second equation from the first yields $$\frac1x - \frac1z = \frac{15}{42}$$
and adding that to the last
$$\frac{2}{x} = \frac{42}{42} = 1 \Rightarrow x = 2.$$
Subtracting from the last yields
$$\frac{2}{z} = \frac{12}{42} \Rightarrow z = 7.$$
Inserting $x = 2$ for example into the first equation yields $y = 3$.
The formula for $x$ and $z$ in the link is
$$ x = z - \frac{p}{3z} - \frac{b}{3a} \qquad \text{and}\qquad z = \sqrt[3]{-\frac{q}{2} \pm \sqrt{D}}, $$
where $p$, $q$, and $D$ are defined by
$$ p = -\frac{b^2}{3a^2} + \frac{c}{a}, \qquad q = \frac{2b^3}{27a^3} - \frac{bc}{3a^2} + \frac{d}{a}, \qquad\text{and}\qquad D = \frac{q^2}{4} + \frac{p^3}{27}. $$
Applying this to $-2x^3+3x^2-x+5=0$, we obtain
$$ p = -0.25, \qquad q = -2.5, \qquad D \approx 1.56192. $$
Then the six possible values of $z$, denoted by
$$ z_{k,\pm} = e^{2k\pi i/3} \sqrt[3]{-\frac{q}{2} \pm \sqrt{D}}, $$
are given by
\begin{align*}
z_{0,+} &\approx 1.35717 & \Rightarrow \quad x &\approx 1.91857, \\
z_{1,+} &\approx -0.678583 + 1.17534 i & \Rightarrow \quad x &\approx -0.209285 + 1.12216 i, \\
z_{2,+} &\approx -0.678583 - 1.17534 i & \Rightarrow \quad x &\approx -0.209285 - 1.12216 i, \\
z_{0,-} &\approx 0.0614024 & \Rightarrow \quad x &\approx 1.91857, \\
z_{1,-} &\approx -0.0307012 + 0.0531761 i & \Rightarrow \quad x &\approx -0.209285 - 1.12216 i, \\
z_{2,-} &\approx -0.0307012 - 0.0531761 i & \Rightarrow \quad x &\approx -0.209285 + 1.12216 i. \\
\end{align*}
So I suspect that you made some mistakes. Note that, when computing $z_{k,-}$'s, you have to work with the expression
$$ -\frac{q}{2} - \sqrt{D} $$
where
$$-\frac{q}{2} = 1.25 \qquad\text{and}\qquad \sqrt{D} \approx 1.2497684970810779307.$$
Since these values are very close, their difference leads to the loss of several significant digits. For instance, if we use six digits, then
$$ -\frac{q}{2} - \sqrt{D} \approx (1.25) - (1.24977) = 0.00023, $$
losing four digits in the process! Now, given that you are working under only three significant digits, you will almost certainly lose all the significant digits in this process, ending up with quantities dominated by rounding errors. I strongly suspect that this is the source of your incorrect answer.
Best Answer
The polynomial from part (a) is
$$P(t) : t^3 - 2t^2 -123t + 540 = 0$$
Now, factorizing
$$P(t) = (t-5)(t-9)(t+12) = 0$$
Since it has no repeated roots, the original equation has distinct solutions corresponding to the following 6 match-ups
and so on.. check for duplicates