If the mirror image of $P(a,6,9)$ with respect to line $\frac{x-3}{7} =\frac{y-2}{5}=\frac{z-1}{-9}$..

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If the mirror image of $P(a,6,9)$ with respect to line $\frac{x-3}{7} =\frac{y-2}{5}=\frac{z-1}{-9}$ is $(20,b,-a-9)$, then find $|a+b|$

The general point of the line is $(3+7k,2+5k,1-9k)$

The direction ratio line joining this point to P is $$(3+7k-a,5k-4, -8-9k)$$

So the dot product of the two direction ratios must be zero

So $$k=\frac{7a-73}{155}$$

Now using the fact that $$3+7k=\frac{20+a}{2}$$

Gives $a=\frac{-3192}{57}$

Beyond which all the values turn into a random mess. Now I don’t know whether I am making computation error or not (even though I tried it many times), or whether my concept is wrong. But I am not getting the answer with this. Where am I going wrong?

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For some $k$, $(3 + 7k, 2 + 5k, 1-9k)$ is the mid-point of $(a,6,9)$ and $(20, b, -a-9$), hence we have:

\begin{cases} \dfrac {20+a}2 &= 3+7k\\ \dfrac {b+6}2 &= 2 + 5k\\ \dfrac {9-a-9}2 &= 1-9k \end{cases}

Solving the system above gives $a=-56, b=-32, k = -3$.