To find the length of the indicated portion of the trajectory, we'll integrate the magnitude of the velocity vector over the given interval \([-1, 0]\). The velocity vector \(\mathbf{v}(t)\) is the derivative of the position vector \(\mathbf{r}(t)\):
\[\mathbf{v}(t) = \frac{d\mathbf{r}}{dt}\]
Given \(\mathbf{r}(t) = (5+2t)\mathbf{i} + (4+3t)\mathbf{j} + (5-6t)\mathbf{k}\), we can find \(\mathbf{v}(t)\) by differentiating each component with respect to \(t\):
\[\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \frac{d}{dt}[(5+2t)\mathbf{i} + (4+3t)\mathbf{j} + (5-6t)\mathbf{k}]\]
Now, differentiate each component separately:
\[\begin{align*}
\frac{d}{dt}(5+2t)\mathbf{i} &= 2\mathbf{i} \\
\frac{d}{dt}(4+3t)\mathbf{j} &= 3\mathbf{j} \\
\frac{d}{dt}(5-6t)\mathbf{k} &= -6\mathbf{k}
\end{align*}\]
So, the velocity vector \(\mathbf{v}(t)\) is:
\[\mathbf{v}(t) = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\]
Now, we want to find the magnitude of \(\mathbf{v}(t)\) and then integrate it over the given interval \([-1, 0]\):
\[\begin{align*}
|\mathbf{v}(t)| &= \sqrt{(2)^2 + (3)^2 + (-6)^2} \\
|\mathbf{v}(t)| &= \sqrt{4 + 9 + 36} \\
|\mathbf{v}(t)| &= \sqrt{49} \\
|\mathbf{v}(t)| &= 7
\end{align*}\]
Now, we'll integrate \(|\mathbf{v}(t)|\) with respect to \(t\) over the interval \([-1, 0]\):
\[\int_{-1}^{0} 7 \, dt = 7t \bigg|_{-1}^{0} = 7(0) - 7(-1) = 7\]
So, the length of the indicated portion of the trajectory is \(7\). Therefore, the answer is A. 7.