To solve this problem, we need to find the flow line (or integral curve) of the vector field $\vec{F}(x, y, z) = \langle -3x, y^2, 5 \rangle$ that passes through the point $\vec{r}(1) = \langle 2, -1, 3 \rangle$.

Step 1: Set Up the System of Differential Equations

The flow line $\vec{r}(t) = \langle x(t), y(t), z(t) \rangle$ is found by solving the system of differential equations given by:

$\frac{dx}{dt} = -3x, \quad \frac{dy}{dt} = y^2, \quad \frac{dz}{dt} = 5$

Step 2: Solve Each Differential Equation Separately

1. Solve $\frac{dx}{dt} = -3x$

This is a separable differential equation:

$\frac{dx}{x} = -3 dt$

Integrating both sides:

$\ln |x| = -3t + C_1$

Exponentiating both sides:

$x(t) = C_1 e^{-3t}$

2. Solve $\frac{dy}{dt} = y^2$

Rewrite as:

$\frac{dy}{y^2} = dt$

Integrate both sides:

$-\frac{1}{y} = t + C_2$

Solve for $y$:

$y(t) = -\frac{1}{t + C_2}$

3. Solve $\frac{dz}{dt} = 5$

Integrate:

$z(t) = 5t + C_3$

Step 3: Apply the Initial Condition $\vec{r}(1) = \langle 2, -1, 3 \rangle$

Plug $t = 1$ into each equation:

1. For $x(1) = 2$:

   $2 = C_1 e^{-3 \cdot 1} \implies C_1 = 2e^3$

   So, $x(t) = 2e^3 e^{-3t} = 2e^{3(1-t)}$.

2. For $y(1) = -1$:

   $-1 = -\frac{1}{1 + C_2} \implies C_2 = 0$

   So, $y(t) = -\frac{1}{t}$.

3. For $z(1) = 3$:

   $3 = 5 \cdot 1 + C_3 \implies C_3 = -2$

   So, $z(t) = 5t - 2$.

Step 4: Write the Flow Line $\vec{r}(t)$

The flow line is:

$\vec{r}(t) = \langle 2e^{3(1-t)}, -\frac{1}{t}, 5t - 2 \rangle.$

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. How do we solve differential equations of different types (separable, linear, etc.)?
2. How can initial conditions affect the solution of a differential equation?
3. What are flow lines and how do they relate to vector fields?
4. How do you find the direction of flow lines from a given vector field?
5. How does the exponential function behave as $t$ approaches infinity or zero?

Tip: When solving differential equations, always verify your solution by substituting it back into the original equation to ensure consistency.