Problem 71

Let's approach the problem step-by-step, maintaining close adherence to the given solution:

1. **Identify the coordinates of point $A$ and point $B$:**
- Point $A$ is given as $(-1, 2, 0)$.
- Point $B$ is given as $(-1, 1, 2)$.

2. **Find the coordinates of point $C$ given it is symmetric to $B$ across the $y$-axis:**
To find a point symmetric to $B$ across the $y$-axis, we negate the $x$-coordinate and keep the $y$ and $z$ coordinates the same. Thus,
- Point $C$ has coordinates $(1, 1, -2)$.

3. **Calculate the distance $|\overrightarrow{AC}|$:**
The formula for the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three-dimensional space is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$. Applying this formula to points $A$ and $C$:
\begin{align*}
|\overrightarrow{AC}| &= \sqrt{(1 - (-1))^2 + (1 - 2)^2 + (-2 - 0)^2} \\
&= \sqrt{(1 + 1)^2 + (1 - 2)^2 + (-2 - 0)^2} \\
&= \sqrt{2^2 + (-1)^2 + (-2)^2} \\
&= \sqrt{4 + 1 + 4} \\
&= \sqrt{9} \\
&= 3.
\end{align*}

Hence, the magnitude of vector $\overrightarrow{AC}$ is $\boxed{3}$.