1. **State the problem:** We have the system of differential equations: $$\frac{dx}{dt} = -x(t) + 3y(t)$$ $$\frac{dy}{dt} = 3y(t) + 5z(t)$$ $$\frac{dz}{dt} = -10z(t)$$ with initial conditions: $$x(0) = 0, \quad y(0) = 0, \quad z(0) = z_0.$$ We need to find explicit expressions for $z(t)$, $y(t)$, and $x(t)$. 2. **Solve for $z(t)$:** The equation for $z(t)$ is independent: $$\frac{dz}{dt} = -10z(t)$$ This is a first-order linear ODE with solution: $$z(t) = z_0 e^{-10t}.$$ 3. **Solve for $y(t)$:** The equation for $y(t)$ is: $$\frac{dy}{dt} = 3y(t) + 5z(t) = 3y(t) + 5 z_0 e^{-10t}.$$ This is a nonhomogeneous linear ODE. The integrating factor is: $$\mu(t) = e^{-3t}.$$ Multiply both sides: $$e^{-3t} \frac{dy}{dt} - 3 e^{-3t} y = 5 z_0 e^{-13t}.$$ Rewrite left side as derivative: $$\frac{d}{dt} \left(y e^{-3t}\right) = 5 z_0 e^{-13t}.$$ Integrate both sides: $$y e^{-3t} = 5 z_0 \int e^{-13t} dt + C = 5 z_0 \left(-\frac{1}{13} e^{-13t}\right) + C.$$ Apply initial condition $y(0) = 0$: $$0 = y(0) = e^{0} y(0) = 5 z_0 \left(-\frac{1}{13} e^{0}\right) + C \Rightarrow C = \frac{5}{13} z_0.$$ So: $$y e^{-3t} = -\frac{5}{13} z_0 e^{-13t} + \frac{5}{13} z_0,$$ which gives $$y(t) = \frac{5}{13} z_0 \left(1 - e^{-10t}\right) e^{3t} = \frac{5}{13} z_0 \left(e^{3t} - e^{-10t}\right).$$ 4. **Solve for $x(t)$:** The equation for $x(t)$ is: $$\frac{dx}{dt} = -x(t) + 3 y(t).$$ Rewrite as: $$\frac{dx}{dt} + x = 3 y(t).$$ The integrating factor is: $$e^{t}.$$ Multiply both sides: $$e^{t} \frac{dx}{dt} + e^{t} x = 3 e^{t} y(t) \Rightarrow \frac{d}{dt} \left(x e^{t}\right) = 3 e^{t} y(t).$$ Substitute $y(t)$: $$\frac{d}{dt} \left(x e^{t}\right) = 3 e^{t} \cdot \frac{5}{13} z_0 \left(e^{3t} - e^{-10t}\right) = \frac{15}{13} z_0 \left(e^{4t} - e^{-9t}\right).$$ Integrate both sides: $$x e^{t} = \frac{15}{13} z_0 \int \left(e^{4t} - e^{-9t}\right) dt + C = \frac{15}{13} z_0 \left(\frac{1}{4} e^{4t} + \frac{1}{9} e^{-9t}\right) + C.$$ Apply initial condition $x(0) = 0$: $$0 = x(0) = e^{0} x(0) = \frac{15}{13} z_0 \left(\frac{1}{4} + \frac{1}{9}\right) + C = \frac{15}{13} z_0 \cdot \frac{13}{36} + C = \frac{15}{36} z_0 + C,$$ so $$C = -\frac{15}{36} z_0 = -\frac{5}{12} z_0.$$ Therefore: $$x e^{t} = \frac{15}{13} z_0 \left(\frac{1}{4} e^{4t} + \frac{1}{9} e^{-9t}\right) - \frac{5}{12} z_0,$$ which simplifies to $$x(t) = z_0 \left(\frac{15}{52} e^{3t} + \frac{5}{39} e^{-10t} - \frac{5}{12} e^{-t}\right).$$ 5. **Compare with given options:** Rewrite $x(t)$ as: $$x(t) = \frac{5}{156} z_0 \left(9 e^{3t} + 4 e^{-10t} - 13 e^{-t}\right),$$ which matches the first and last options for $x(t)$. Similarly, $y(t) = \frac{5}{13} z_0 \left(e^{3t} - e^{-10t}\right)$ matches the first option. Hence, the correct choice is: $$x(t) = \frac{5}{156} z_0 \left(4 e^{-10t} - 13 e^{-t} + 9 e^{3t}\right), \quad y(t) = \frac{5}{13} z_0 \left(- e^{-10t} + e^{3t}\right), \quad z(t) = z_0 e^{-10t}.$$