The minimum possible value of $$\frac{x^{2} - 6x + 10}{3-x}$$, for $$x < 3$$, is

Let $$\frac{x^2-6x+10}{3-x}=p$$

$$x^2-6x+10=3p-px$$

$$x^2-\left(6-p\right)x+10-3p=0$$

Since the equation will have real roots,

$$\left(6-p\right)^2-4\times\left(10-3p\right)\ge0$$

$$p^2-12p+12p+36-40\ge0$$

$$p^2\ge4$$

$$p\ge2\ ,\ p\le-2$$

Now, when $$p=-2$$, $$x = 4$$.  Since it is given that $$x<3$$, thus this value will be discarded.

Now, $$\frac{1}{2}$$ and $$-\frac{1}{2}$$ do not come in the mentioned range.

when $$p=2$$, $$x = 2$$

Thus, the minimum possible value of p will be 2.

Thus, the correct option is B.

Alternate explanation:

Since $$x<3$$,

$$3-x>0$$

Let $$3-x=y$$. So, $$y>0$$.

Now, $$\frac{x^2-6x+10}{3-x}=\frac{x^2-6x+9+1}{3-x}$$

=> $$\frac{\left(3-x\right)^2+1}{3-x}$$

Since $$3-x=y$$, the equation will transform to $$\frac{y^2+1}{y}$$ or $$y+\frac{1}{y}$$

The minimum value of the expression $$y+\frac{1}{y}$$ for $$y>0$$ will at $$y=1$$

i.e., Minimum value = $$1+1=2$$

Thus, the correct option is B.