To find the equation and length of the altitude from vertex A in triangle ABC with vertices A(2,3), B(4,-1), and C(1,2), we first determine the equation of side BC. The slope of BC is calculated as:

$$m_{BC} = \frac{2 - (-1)}{1 - 4} = \frac{3}{-3} = -1$$

Using point B(4,-1), the equation of BC is:

$$y - (-1) = -1(x - 4)$$

which simplifies to:

$$x + y - 3 = 0$$

The altitude from A is perpendicular to BC, so its slope is the negative reciprocal of $-1$, i.e., $m = 1$. Since it passes through A(2,3), its equation is:

$$y - 3 = 1(x - 2)$$

giving:

$$x - y + 1 = 0$$

For the length of the altitude, we use the distance from point A to line BC. The distance formula gives:

$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Substituting values:

$$d = \frac{|2 + 3 - 3|}{\sqrt{1+1}} = \frac{2}{\sqrt{2}} = \sqrt{2} \text{ units}$$

Alternatively, using the area of the triangle:

$$\text{area} = \frac{1}{2} |2(-1-2) + 4(2-3) + 1(3-(-1))| = \frac{1}{2} | -6 -4 +4| = 3$$

The length of BC is:

$$\sqrt{(4-1)^2 + (-1-2)^2} = \sqrt{9+9} = 3\sqrt{2}$$

Hence, altitude $= \frac{2 \times \text{area}}{BC} = \frac{6}{3\sqrt{2}} = \sqrt{2}$. Both methods confirm the altitude length as $\sqrt{2}$.

Thus, the equation of the altitude is $x - y + 1 = 0$ and its length is $\sqrt{2}$ units.