1. **Problem statement:** We need to find the line integral of the scalar function $$f(x,y,z) = \frac{\sqrt{3}}{x^2 + y^2 + z^2}$$ along the curve defined by $$\mathbf{r}(t) = t\mathbf{i} + t\mathbf{j} + t\mathbf{k}$$ for $$0 \leq t < \infty$$. 2. **Parameterization and formula:** The curve is parameterized by $$\mathbf{r}(t) = (t, t, t)$$. The line integral of a scalar function along a curve is given by: $$\int_C f(x,y,z) \, ds = \int_a^b f(\mathbf{r}(t)) \|\mathbf{r}'(t)\| \, dt$$ where $$\mathbf{r}'(t)$$ is the derivative of $$\mathbf{r}(t)$$ and $$\|\mathbf{r}'(t)\|$$ is its magnitude. 3. **Calculate $$\mathbf{r}'(t)$$ and its magnitude:** $$\mathbf{r}'(t) = \frac{d}{dt}(t, t, t) = (1, 1, 1)$$ $$\|\mathbf{r}'(t)\| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$$ 4. **Substitute into the integral:** Since $$x = t, y = t, z = t$$, we have: $$f(\mathbf{r}(t)) = \frac{\sqrt{3}}{t^2 + t^2 + t^2} = \frac{\sqrt{3}}{3t^2} = \frac{\sqrt{3}}{3t^2}$$ The integral becomes: $$\int_0^\infty \frac{\sqrt{3}}{3t^2} \times \sqrt{3} \, dt = \int_0^\infty \frac{3}{3t^2} \, dt = \int_0^\infty \frac{1}{t^2} \, dt$$ 5. **Evaluate the integral:** $$\int_0^\infty \frac{1}{t^2} \, dt = \lim_{a \to 0^+} \int_a^1 t^{-2} \, dt + \int_1^\infty t^{-2} \, dt$$ Calculate each part: - $$\int_a^1 t^{-2} \, dt = \left[-\frac{1}{t}\right]_a^1 = 1 - \frac{1}{a}$$ which diverges as $$a \to 0^+$$. - $$\int_1^\infty t^{-2} \, dt = \left[-\frac{1}{t}\right]_1^\infty = 1$$ Since the integral near zero diverges, the entire integral diverges. **Final answer:** The line integral diverges (does not converge to a finite value).