If the line $$l_1: 3y - 2x = 3$$ is the angular bisector of the lines $$l_2: x - y + 1 = 0$$ and $$l_3: \alpha x + \beta y + 17 = 0$$, then $$\alpha^2 + \beta^2 - \alpha - \beta$$ is equal to _______

Given $$l_1: 3y - 2x = 3$$ is the angular bisector of $$l_2: x - y + 1 = 0$$ and $$l_3: \alpha x + \beta y + 17 = 0$$.

Finding the intersection of $$l_1$$ and $$l_2$$:

From $$l_2$$: $$x = y - 1$$. Substituting into $$l_1$$: $$3y - 2(y-1) = 3 \implies y = 1, x = 0$$.

Intersection point: $$(0, 1)$$.

Since $$l_1$$ bisects the angle between $$l_2$$ and $$l_3$$, and the bisector passes through the intersection of $$l_2$$ and $$l_3$$, the point $$(0, 1)$$ lies on $$l_3$$:

$$\beta + 17 = 0 \implies \beta = -17$$

Using the equal angle condition:

Slope of $$l_1 = \frac{2}{3}$$, slope of $$l_2 = 1$$, slope of $$l_3 = \frac{\alpha}{17}$$.

Angle between $$l_2$$ and $$l_1$$:

$$\tan\theta_1 = \left|\frac{1 - 2/3}{1 + 2/3}\right| = \frac{1/3}{5/3} = \frac{1}{5}$$

Angle between $$l_3$$ and $$l_1$$:

$$\tan\theta_2 = \left|\frac{\alpha/17 - 2/3}{1 + 2\alpha/51}\right| = \frac{|3\alpha - 34|}{|51 + 2\alpha|}$$

Setting $$\tan\theta_1 = \tan\theta_2$$:

$$\frac{|3\alpha - 34|}{|51 + 2\alpha|} = \frac{1}{5}$$

Case 1: $$3\alpha - 34 = \frac{51 + 2\alpha}{5} \implies \alpha = 17$$

But this gives $$l_3: 17x - 17y + 17 = 0 \equiv x - y + 1 = 0 = l_2$$. Invalid.

Case 2: $$3\alpha - 34 = -\frac{51 + 2\alpha}{5} \implies 17\alpha = 119 \implies \alpha = 7$$

Verification: $$l_3: 7x - 17y + 17 = 0$$. The bisectors of $$l_2$$ and $$l_3$$ include $$-2x + 3y - 3 = 0$$, which is $$l_1$$. ✓

Result:

$$\alpha^2 + \beta^2 - \alpha - \beta = 49 + 289 - 7 - (-17) = 49 + 289 - 7 + 17 = 348$$

The answer is $$348$$.