If $$(a, b, c)$$ is the image of the point $$(1, 2, -3)$$ in the line, $$\frac{x+1}{2} = \frac{y-3}{-2} = \frac{z}{-1}$$, then $$a + b + c$$ is equal to:

We have to find the image of the point $$P(1,\,2,\,-3)$$ in the given line

$$\frac{x+1}{2} \;=\; \frac{y-3}{-2} \;=\; \frac{z}{-1}$$

Let us denote the image by $$I(a,\,b,\,c)$$. Any point on the line can be written in the symmetric-to-parametric form by introducing a parameter, say $$\lambda$$. First we rewrite the line:

$$\frac{x+1}{2}= \frac{y-3}{-2}= \frac{z}{-1}= \lambda.$$

So a general point on the line is

$$x = -1 + 2\lambda,\qquad y = 3 - 2\lambda,\qquad z = 0 - 1\lambda = -\lambda.$$

Hence we may set

$$I(a,\,b,\,c)\;=\;\bigl(-1 + 2\lambda,\; 3 - 2\lambda,\; -\lambda\bigr).$$

For $$I$$ to be the image of $$P$$ in the line, the segment $$PI$$ must be perpendicular to the line. The direction ratios of the line are read directly from its symmetric form and are

$$\vec{d} = \langle 2,\,-2,\,-1\rangle.$$

The vector joining $$I$$ to $$P$$ is

$$\vec{PI} = \langle\,1 - (-1 + 2\lambda),\; 2 - (3 - 2\lambda),\; -3 - (-\lambda)\rangle = \langle\,2 - 2\lambda,\; -1 + 2\lambda,\; -3 + \lambda\rangle.$$

Perpendicularity condition: the dot-product of $$\vec{PI}$$ with the direction vector $$\vec{d}$$ must be zero.

$$\vec{PI}\,\cdot\,\vec{d} \;=\; (2 - 2\lambda)(2) \;+\; (-1 + 2\lambda)(-2) \;+\; (-3 + \lambda)(-1)\;=\;0.$$

We expand every term carefully:

$$\begin{aligned} (2 - 2\lambda)(2) &= 4 - 4\lambda,\\[2pt] (-1 + 2\lambda)(-2) &= 2 - 4\lambda,\\[2pt] (-3 + \lambda)(-1) &= 3 - \lambda. \end{aligned}$$

Adding them,

$$\bigl(4 - 4\lambda\bigr) + \bigl(2 - 4\lambda\bigr) + \bigl(3 - \lambda\bigr) = 4 + 2 + 3 \;-\; (4\lambda + 4\lambda + \lambda) = 9 - 9\lambda.$$

Setting this equal to zero,

$$9 - 9\lambda = 0 \;\Longrightarrow\; \lambda = 1.$$

Substituting $$\lambda = 1$$ back into the coordinates of $$I$$ we get

$$a = -1 + 2(1) = 1,\qquad b = 3 - 2(1) = 1,\qquad c = -1.$$

Therefore,

$$a + b + c = 1 + 1 - 1 = 1.$$

Hence, the correct answer is Option D.