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 $\mathit{a}\mathbf{+}\mathit{b}\mathbf{+}\mathit{c}$ is

The correct option is A

$2$

Finding the value :

Given, $(a,b,c)$is the image of the point $(1,2,-3)$

Line is $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}=\lambda$ Let point $M$is $(2\lambda -1,-2\lambda +3,-\lambda )$

Direction ration of $PP\text{'}\equiv (2\lambda -2,-2\lambda +1,3-\lambda )$

PM is Perpendicular to line

$\begin{array}{rcl}2(2\lambda -2)-2(-2\lambda +1)-1(3-\lambda )& =& 0\\ & \Rightarrow & 4\lambda -4+4\lambda -2-3+\lambda =0\\ \Rightarrow 9\lambda & =& 9\\ \Rightarrow \lambda & =& 1\end{array}$

Point $M$ is $(1,1,-1)$

$\begin{array}{rcl}\frac{a+1}{2}& =& 1\\ & \Rightarrow & a=1\end{array}$,$\begin{array}{rcl}\frac{b+2}{2}& =& 1\\ & \Rightarrow & b=0\end{array}$, $\begin{array}{rcl}\frac{c-3}{2}& =& -1\\ & \Rightarrow & c=1\end{array}$

Therefore, $\mathit{a}\mathbf{+}\mathit{b}\mathbf{+}\mathit{c}\mathbf{=}\mathbf{2}$

Hence, correct option is $\mathbf{\left(}\mathit{A}\mathbf{\right)}$