Let $$Q$$ be the mirror image of the point $$P(1, 2, 1)$$ with respect to the plane $$x + 2y + 2z = 16$$. Let $$T$$ be a plane passing through the point $$Q$$ and contains the line $$\vec{r} = -\hat{k} + \lambda(\hat{i} + \hat{j} + 2\hat{k})$$, $$\lambda \in R$$. Then, which of the following points lies on $$T$$?

The plane is

$$x+2y+2z=16$$

and the point is

$$P(1,2,1)$$

Using the image formula,

$$Q=P-2\left(\frac{ax_0+by_0+cz_0+d}{a^2+b^2+c^2}\right)(a,b,c)$$

Here,

$$a=1,\ b=2,\ c=2,\ d=-16$$

So,

$$ax_0+by_0+cz_0+d=1+4+2-16=-9$$

and

$$a^2+b^2+c^2=1+4+4=9$$

Hence,

$$Q=(1,2,1)-2\left(\frac{-9}{9}\right)(1,2,2)$$

$$Q=(1,2,1)+2(1,2,2)$$

$$Q=(3,6,5)$$

The given line is

$$\vec r=-\hat k+\lambda(\hat i+\hat j+2\hat k)$$

So, a point on the line is

$$A(0,0,-1)$$

and its direction vector is

$$\vec d=(1,1,2)$$

Now,

$$\overrightarrow{AQ}=(3,6,6)$$

A normal vector to plane $$T$$ is

$$\vec n=\vec d\times \overrightarrow{AQ}$$

$$\vec n=(-6,0,3)$$

$$\vec n=(-2,0,1)$$

Hence, equation of plane $$T$$ is

$$-2(x-3)+(z-5)=0$$

$$2x-z-1=0$$

Checking the options:

For $$(1,2,1)$$,

$$2(1)-1-1=0$$

Hence, $$(1,2,1)$$ lies on the plane.

Therefore, the correct answer is

$$\boxed{\text{Option B }(1,2,1)}$$