$A$ line $L$ is passing through points $A(1,3,2)$ and $B(2,2,1)$. If mirror image of point $\mathrm{P}(1,1…

Direction vector of the line through $A(1,3,2)$ and $B(2,2,1)$ is

$\vec{d} = \vec{AB} = (2-1, 2-3, 1-2) = (1,-1,-1)$

Parametric coordinates of any point on the line are

$M = (1+t, 3-t, 2-t)$

For the perpendicular foot from $P(1,1,-1)$, vector $\vec{PM} = (t, 2-t, 3-t)$ must satisfy $\vec{PM} \cdot \vec{d} = 0$:

$t(1) + (2-t)(-1) + (3-t)(-1) = 0$

$t - 2 + t - 3 + t = 0$

$3t = 5$

$t = \frac{5}{3}$

Substituting gives the foot coordinates

$M = \left(1+\frac{5}{3}, 3-\frac{5}{3}, 2-\frac{5}{3}\right) = \left(\frac{8}{3}, \frac{4}{3}, \frac{1}{3}\right)$

Since $M$ is the midpoint of $P$ and its mirror image $P'(x,y,z)$, we solve:

$\frac{1+x}{2} = \frac{8}{3} \Rightarrow x = \frac{13}{3}$

$\frac{1+y}{2} = \frac{4}{3} \Rightarrow y = \frac{5}{3}$

$\frac{-1+z}{2} = \frac{1}{3} \Rightarrow z = \frac{5}{3}$

$x+y+z = \frac{13}{3} + \frac{5}{3} + \frac{5}{3} = \frac{23}{3}$

Final answer: $\boxed{\frac{23}{3}}$