Let $$S$$ be the reflection of a point $$Q$$ with respect to the plane given by

$$\vec{r} = -(t+p)\hat{i} + t\hat{j} + (1+p)\hat{k}$$

where $$t$$, $$p$$ are real parameters and $$\hat{i}$$, $$\hat{j}$$, $$\hat{k}$$ are the unit vectors along the three positive coordinate axes. If the position vectors of $$Q$$ and $$S$$ are $$10\hat{i} + 15\hat{j} + 20\hat{k}$$ and $$\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$$ respectively, then which of the following is/are TRUE?

The given parametric representation contains two free parameters $$t$$ and $$p$$, therefore it represents a plane.
$$\vec r = -(t+p)\hat i + t\hat j + (1+p)\hat k$$

Direction vector obtained by varying $$t$$ while keeping $$p$$ fixed:
$$\vec v_1 = \frac{\partial\vec r}{\partial t}=(-1,\,1,\,0)$$

Direction vector obtained by varying $$p$$ while keeping $$t$$ fixed:
$$\vec v_2 = \frac{\partial\vec r}{\partial p}=(-1,\,0,\,1)$$

The normal vector of the plane is the cross product $$\vec n=\vec v_1\times\vec v_2$$:
$$\vec n =\begin{vmatrix} \hat i & \hat j & \hat k\\ -1 & 1 & 0\\ -1 & 0 & 1 \end{vmatrix} = (1,\,1,\,1)$$

Hence the cartesian equation of the plane is of the form $$x+y+z=d$$. Substituting any point on the plane, say for $$t=0,\;p=0$$ we get $$\vec r_0=(0,0,1)$$, thus
$$0+0+1=d\quad\Longrightarrow\quad d=1$$

Therefore, the required plane is
$$x+y+z=1$$

Let the position vector of $$Q$$ be $$\vec Q=(10,15,20)$$ and that of its reflection $$S$$ be $$\vec S=(\alpha,\beta,\gamma)$$. For a plane $$\vec n\cdot\vec r=1$$ with normal $$\vec n=(1,1,1)$$, the reflection formula is

$$\vec S=\vec Q-2\,\frac{\vec n\cdot\vec Q-1}{\lVert\vec n\rVert^{2}}\;\vec n$$

Compute the required quantities:
$$\vec n\cdot\vec Q = 1\cdot10+1\cdot15+1\cdot20 = 45$$
$$\vec n\cdot\vec Q - 1 = 44$$
$$\lVert\vec n\rVert^{2}=1^{2}+1^{2}+1^{2}=3$$

Hence
$$\vec S=(10,15,20)-2\left(\frac{44}{3}\right)(1,1,1) =(10,15,20)-\left(\frac{88}{3},\frac{88}{3},\frac{88}{3}\right)$$

So
$$\alpha = 10-\frac{88}{3}=\frac{-58}{3},\quad \beta = 15-\frac{88}{3}=\frac{-43}{3},\quad \gamma = 20-\frac{88}{3}=\frac{-28}{3}$$

Now test each option:

$$3(\alpha+\beta)=3\left(\frac{-58}{3}+\frac{-43}{3}\right)=3\left(\frac{-101}{3}\right)=-101$$ Option A is true.

$$3(\beta+\gamma)=3\left(\frac{-43}{3}+\frac{-28}{3}\right)=3\left(\frac{-71}{3}\right)=-71$$ Option B is true.

$$3(\gamma+\alpha)=3\left(\frac{-28}{3}+\frac{-58}{3}\right)=3\left(\frac{-86}{3}\right)=-86$$ Option C is true.

$$3(\alpha+\beta+\gamma)=3\left(\frac{-58-43-28}{3}\right)=3\left(\frac{-129}{3}\right)=-129$$ This does not match the value given in Option D, so Option D is false.

Therefore, the correct statements are:
Option A, Option B, and Option C.