Reflection over a line

Let the line be $\vec{\ell}(t)=\vec{a}+t\hat{v}$ , where $\vec{a}$ is one initial point, $\hat{v}$ a unit parallel vector, and $\vec{P}$ the point we will reflect.

We will project the vector $\vec{P}-\vec{a}$ in the line and call it $\vec{u}$ : $\vec{u}=\left(\left(\vec{P}-\vec{a}\right) \cdot \hat{v}\right)\hat{v}$ Now let $\vec{d}$ the perpendicular vector from $\vec{P}$ to the line: $\vec{d}=\vec{u}-\left(\vec{P}-\vec{a}\right)=\vec{u}+\vec{a}-\vec{P}$ So, the new point $\vec{P'}$ can be obtained if we add two times the vector $\vec{d}$ to the point $\vec{P}$ : $\vec{P'} = \vec{P} + 2\vec{d}$ Finally, putting all together: $\vec{P'} = \vec{P} + 2\left(\vec{u}+\vec{a}-\vec{P}\right) \\ \vec{P'} = 2\vec{a}-\vec{P}+2\left(\left(\vec{P}-\vec{a}\right) \cdot \hat{v}\right)\hat{v}$ In this case $\vec{a}=(3,4,1)$ , $\hat{v}=\dfrac{1}{3}(1,2,2)$ and $\vec{P}=(-3,5,2)$ , so the new point will be $\vec{P'}=\dfrac{1}{9}(77,19,-8)$ , making the answer $\boxed{88}$ .

Let $P(k+3,2k+4,2k+1)$ be the foot of perpendicular on the given line from point $A(-3,5,2)$ and $\vec{v}=\hat{i}+2\hat{j}+2\hat{k}$ be a vector parallel to the line.

$\vec{AP}.\vec{v}=0\Rightarrow k=\frac{-2}{9} \therefore P (\frac{25}{9},\frac{32}{9},\frac{5}{9})$

Now, we can see that $P$ is the mid point of $A$ and its image. Therefore, $(a,b,c)\equiv(\frac{77}{9},\frac{19}{9},-\frac{8}{9})$