LetA(l, 0), B(2, -1) and $$C\left(\frac{7}{3}, \frac{4}{3}\right)$$ be three points. If the equation of the bisector of the angle ABC is $$\alpha x+\beta y=5$$, then the value of $$\alpha^{2} +\beta^{2}$$ is

The vector $$\vec{BA} = A - B = (-1,1)$$ has magnitude $$|\vec{BA}| = \sqrt{2}$$, and the vector $$\vec{BC} = C - B = \bigl(\tfrac{1}{3},\tfrac{7}{3}\bigr)$$ has magnitude $$|\vec{BC}| = \tfrac{1}{3}\sqrt{1+49} = \tfrac{5\sqrt{2}}{3}\,.$$

Using these, the unit vectors are $$\hat{BA} = \frac{(-1,1)}{\sqrt{2}} = \bigl(-\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}\bigr)$$ and $$\hat{BC} = \frac{(1/3,7/3)}{5\sqrt{2}/3} = \frac{(1,7)}{5\sqrt{2}} = \bigl(\tfrac{1}{5\sqrt{2}},\tfrac{7}{5\sqrt{2}}\bigr)\,.$$

Since the angle bisector direction is the sum of these unit vectors, we get$$\hat{BA}+\hat{BC}=\Bigl(-\tfrac{1}{\sqrt{2}}+\tfrac{1}{5\sqrt{2}},\;\tfrac{1}{\sqrt{2}}+\tfrac{7}{5\sqrt{2}}\Bigr)=\Bigl(\tfrac{-4}{5\sqrt{2}},\;\tfrac{12}{5\sqrt{2}}\Bigr)\,,$$which is proportional to $$(-4,12)$$ or simplified to $$(-1,3)\,$$.

Writing the line through $$B(2,-1)$$ in parametric form $$(x,y)=(2,-1)+t(-1,3)$$ and eliminating $$t$$ via $$\frac{x-2}{-1}=\frac{y+1}{3}$$ gives $$3(x-2)=-(y+1)\Rightarrow3x-6=-y-1\Rightarrow3x+y=5\,. $$ Hence $$\alpha=3$$ and $$\beta=1\,. $$

Substituting into $$\alpha^2+\beta^2$$ yields $$9+1=10\,$$, so $$\alpha^2+\beta^2=10\,$$, which matches Option C. Therefore, the answer is Option C.