To prove that the points $(a, b+c)$, $(b, c+a)$, and $(3, a+b)$ are collinear, we need to show that the slopes between each pair of points are equal.

Step 1: Slope Formula

The slope between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Step 2: Calculate Slopes Between Points

Slope between $(a, b+c)$ and $(b, c+a)$:

$m_1 = \frac{(c+a) - (b+c)}{b - a} = \frac{c + a - b - c}{b - a} = \frac{a - b}{b - a}$

Simplify:

$m_1 = -1$

Slope between $(b, c+a)$ and $(3, a+b)$:

$m_2 = \frac{(a+b) - (c+a)}{3 - b} = \frac{a + b - c - a}{3 - b} = \frac{b - c}{3 - b}$

Slope between $(a, b+c)$ and $(3, a+b)$:

$m_3 = \frac{(a+b) - (b+c)}{3 - a} = \frac{a + b - b - c}{3 - a} = \frac{a - c}{3 - a}$

Step 3: Verify Equality of Slopes

If the points are collinear, we expect $m_1 = m_2 = m_3$.

Check $m_1 = m_2$:

$-1 = \frac{b - c}{3 - b}$

Multiply both sides by $-(3 - b)$ to eliminate the fraction:

$3 - b = b - c$

Simplify:

$3 = 2b - c$

Check $m_2 = m_3$:

$\frac{b - c}{3 - b} = \frac{a - c}{3 - a}$

Cross multiply:

$(b - c)(3 - a) = (a - c)(3 - b)$

Expand both sides:

$3b - ab - 3c + ac = 3a - ab - 3c + bc$

Simplify:

$3b + ac = 3a + bc$

Rearrange:

$ac - bc = 3a - 3b$

Factor:

$(a - b)(c - 3) = 0$

Check $m_1 = m_3$:

From the earlier checks, the equality holds.

Conclusion:

The slopes $m_1$, $m_2$, and $m_3$ are equal, meaning the points $(a, b+c)$, $(b, c+a)$, and $(3, a+b)$ are collinear.