Let $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}=\lambda$ (say)
Any point on the line is $\mathrm{P}(3 \lambda+6,2 \lambda+7,-2 \lambda+7)$
Let $A \equiv(1,2,3)$
The d.r.s. of line AP are
$\begin{aligned}
& 3 \lambda+6-1,2 \lambda+7-2,-2 \lambda+7-3 \\
& \text { i.e. } 3 \lambda+5,2 \lambda+5,-2 \lambda+4
\end{aligned}$
Since AP is perpendicular to the given line,
$\begin{aligned}
& 3(3 \lambda+5)+2(2 \lambda+5)-2(-2 \lambda+4)=0 \\
& \Rightarrow 17 \lambda+17=0 \\
& \Rightarrow \lambda=-1 \\
& \therefore \quad \mathrm{P} \equiv(3,5,9) \\
& \therefore \quad \mathrm{AP}=\sqrt{(3-1)^2+(5-2)^2+(9-3)^2} \\
& \quad=\sqrt{49} \\
& \quad=7 \text { units }
\end{aligned}$