Let $$\alpha, \beta, \gamma, \delta$$ be real numbers such that $$\alpha^2 + \beta^2 + \gamma^2 \neq 0$$ and $$\alpha + \beta = 1$$ Suppose the point (3,2,−1) is the mirror image of the point (1,0,−1) with respect to the plane $$\alpha x + \beta y + \gamma z = \delta$$. Then which of the following statements is/are TRUE?

The mirror image condition gives two facts:
(i) The segment joining the original point $$P(1,0,-1)$$ and its image $$Q(3,2,-1)$$ is perpendicular to the reflecting plane, so the vector $$\overrightarrow{PQ}$$ is parallel to the plane’s normal $$(\alpha,\beta,\gamma)$$.
(ii) The midpoint of $$P$$ and $$Q$$ lies on the plane.

Step 1 Find $$\overrightarrow{PQ}$$:
$$\overrightarrow{PQ}=(3-1,\;2-0,\;-1-(-1))=(2,2,0).$$

Step 2 Express the normal vector:
Because $$(\alpha,\beta,\gamma)\parallel(2,2,0),$$ there exists a non-zero constant $$k$$ such that
$$\alpha=2k,\;\beta=2k,\;\gamma=0.$$(This automatically satisfies $$\alpha^2+\beta^2+\gamma^2\neq0$$ for $$k\neq0$$.)

Step 3 Use the given relation $$\alpha+\beta=1$$:
$$2k+2k=1\;\Longrightarrow\;4k=1\;\Longrightarrow\;k=\tfrac14.$$ Hence
$$\alpha=\tfrac12,\qquad\beta=\tfrac12,\qquad\gamma=0.$$

Step 4 Write the plane and determine $$\delta$$ using the midpoint.
With the above values the plane is
$$\tfrac12x+\tfrac12y+0\cdot z=\delta\;\Longrightarrow\;x+y=2\delta.$$ The midpoint $$M$$ of $$P$$ and $$Q$$ is
$$M\Bigl(\tfrac{1+3}{2},\;\tfrac{0+2}{2},\;\tfrac{-1+(-1)}{2}\Bigr)=(2,1,-1).$$ Substituting $$M$$ into the plane:
$$2+1=2\delta\;\Longrightarrow\;3=2\delta\;\Longrightarrow\;\delta=\tfrac32.$$

Step 5 Check each statement.

Option A: $$\alpha+\beta=2.$$
Here $$\alpha+\beta=\tfrac12+\tfrac12=1\neq2.$$ False.

Option B: $$\delta-\gamma=3.$$
$$\delta-\gamma=\tfrac32-0=\tfrac32\neq3.$$ False.

Option C: $$\delta+\beta=4.$$
$$\delta+\beta=\tfrac32+\tfrac12=2\neq4.$$ False.

Option D: $$\alpha+\beta+\gamma=\delta.$$
Left side $$=1+0=1,\;\delta=\tfrac32.$$ Not equal. False.

All four given statements are false. Therefore the correct choice is the “None of these” option, i.e. Option E.