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(A) Let $u = \frac{1}{x}, v = \frac{1}{y}, w = \frac{1}{z}$. The equations become:$u + 2v - 3w = 1$ ...$(i)$$2u - 4v + 3w = 1$ ...(ii)$3u + 6v - 6w =

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$5 \beta$

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(A) Let $u = \frac{1}{x}, v = \frac{1}{y}, w = \frac{1}{z}$. The equations become:$u + 2v - 3w = 1$ ...$(i)$$2u - 4v + 3w = 1$ ...(ii)$3u + 6v - 6w = 4$ ...(iii)Adding $(i)$ and (ii): $3u - 2v = 2$ ...(iv)Multiplying (ii) by $2$ and adding to (iii): $(4u - 8v + 6w) + (3u + 6v - 6w) = 2 + 4 \Rightarrow 7u - 2v = 6$ ...$(v)$Subtracting (iv) from $(v)$: $(7u - 2v) - (3u - 2v) = 6 - 2 \Rightarrow 4u = 4 \Rightarrow u = 1$. Since $u = \frac{1}{x}$,we have $x = 1 = \alpha$.Substituting $u = 1$ into (iv): $3(1) - 2v = 2 \Rightarrow 2v = 1 \Rightarrow v = \frac{1}{2}$. Since $v = \frac{1}{y}$,we have $y = 2 = \beta$.Substituting $u = 1, v = \frac{1}{2}$ into $(i)$: $1 + 2(\frac{1}{2}) - 3w = 1 \Rightarrow 1 + 1 - 3w = 1 \Rightarrow 3w = 1 \Rightarrow w = \frac{1}{3}$. Since $w = \frac{1}{z}$,we have $z = 3 = \gamma$.Thus,$\alpha^2 + \gamma^2 = 1^2 + 3^2 = 1 + 9 = 10$.Since $\beta = 2$,we have $5\beta = 5(2) = 10$.Therefore,$\alpha^2 + \gamma^2 = 5\beta$.

If the solution of the system of simultaneous equations $\frac{1}{x}+\frac{2}{y}-\frac{3}{z}-1=0$,$\frac{2}{x}-\frac{4}{y}+\frac{3}{z}-1=0$ and $\frac{3}{x}+\frac{6}{y}-\frac{6}{z}-4=0$ is $x=\alpha, y=\beta, z=\gamma$,then $\alpha^2+\gamma^2=$

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