Triangle $ABC$ has vertices $A(2, 3),$ $B(1, -8)$ and $C(-3, 2).$ The line containing the altitude through $A$ intersects the point $(0, y).$ What is the value of $y$?
The altitude through $A$ is perpendicular to line $BC.$ First, we find the slope of the line containing $BC,$ which is $\frac{-8 - 3}{1 - (-3)} = \frac{-11}{4}.$ Therefore, the slope of the line containing the altitude through $A$ is $\frac{4}{11}.$ Since this line passes through $(2, 3),$ its equation is given by \[y - 3 = \frac{4}{11}(x - 2).\] Substituting $x = 0,$ we get \[y-3 = -\frac{8}{11} \quad\Rightarrow\quad y = \boxed{\frac{25}{11}}.\]
To find the line containing the altitude through $A,$ we need to first find the slope of $BC$ since the altitude is perpendicular to $BC.$
The slope of $BC$ is given by: $$m_{BC} = \frac{y_2 - y_1}{x_2 - x_1},$$ where $P(x_1, y_1)$ and $Q(x_2, y_2)$ are two points on the line $BC.$
Using the coordinates of $B$ and $C,$ we can calculate the slope of $BC$ as follows:
$$m_{BC} = \frac{(-8) - 2}{1 - (-3)} = \frac{-10}{4} = -\frac{5}{2}.$$
Since the altitude of a triangle is perpendicular to the base, the slope of the line containing the altitude is the negative reciprocal of the slope of $BC.$ Therefore, the slope of the line containing the altitude through $A$ is $\frac{2}{5}.$
Now, we have a point on the line, $(0,y),$ and the slope of the line, which is $\frac{2}{5}.$ Using the point-slope form of a line, we can write the equation of the line:
$$y - y_1 = m(x - x_1),$$ where $m$ is the slope and $(x_1, y_1)$ is a point on the line.
Plugging in the values, we get:
$$y - 3 = \frac{2}{5}(0 - 2).$$
Simplifying, we get:
$$y - 3 = \frac{2}{5}(-2).$$
$$y - 3 = -\frac{4}{5}.$$
To isolate $y,$ we can bring $ -3 $ to the right side of the equation:
$$y = -\frac{4}{5} + 3.$$
$$y = -\frac{4}{5} + \frac{15}{5}.$$
$$y = \frac{11}{5}.$$
Therefore, the value of $y$ is $\boxed{\frac{11}{5}}.$