If $Q(1,0,1)$ is the image of the point $P(a, b, c)$ in the line $\dfrac{x+1}{2}=\dfrac{y-3}{-2}=\dfrac{z}{-1}$ then $a+b+c$ is equal to

Let the line be $L: \dfrac{x+1}{2} = \dfrac{y-3}{-2} = \dfrac{z}{-1} = k$. Any point $M$ on the line is given by $M(2k-1, -2k+3, -k)$.

Since $Q(1, 0, 1)$ is the image of $P(a, b, c)$ in the line, $M$ is the midpoint of $PQ$. Thus, $M = \left( \dfrac{a+1}{2}, \dfrac{b+0}{2}, \dfrac{c+1}{2} \right)$.

Equating the coordinates of $M$ with the parametric form:
$\dfrac{a+1}{2} = 2k-1 \Rightarrow a = 4k-3$
$\dfrac{b}{2} = -2k+3 \Rightarrow b = -4k+6$
$\dfrac{c+1}{2} = -k \Rightarrow c = -2k-1$

The vector $\overrightarrow{PQ} = (1-a, 0-b, 1-c) = (1-(4k-3), -( -4k+6), 1-(-2k-1)) = (4-4k, 4k-6, 2k+2)$ must be perpendicular to the direction vector of the line $\vec{v} = (2, -2, -1)$.

Therefore, $\overrightarrow{PQ} \cdot \vec{v} = 0$:
$2(4-4k) - 2(4k-6) - 1(2k+2) = 0$
$8 - 8k - 8k + 12 - 2k - 2 = 0$
$-18k + 18 = 0 \Rightarrow k = 1$.

Substituting $k=1$ into the expressions for $a, b, c$:
$a = 4(1)-3 = 1$
$b = -4(1)+6 = 2$
$c = -2(1)-1 = -3$

The sum $a+b+c = 1 + 2 - 3 = 0$.

Answer: 0