Equation of the line through the point $P(2,3,-4)$ along the vector $6\hat i+3\hat j-4\hat k$ is
$\large\frac{x-2}{6}=\frac{y-3}{3}=\frac{z+4}{-4}=\lambda$
Any point $Q$ on this line can be given in terms of $\lambda$ as $Q(6\lambda+2,3\lambda+3,-4\lambda-4)$
Let this point $Q$ be the foot of $\perp$ drawn from $A(-1,2,6)$ to the line.
$d.r.$ of $AQ$ is $(6\lambda+3,3\lambda+1,-4\lambda-10)$
Since $AQ$ is $\perp$ to the line, $\overrightarrow {AQ}.(6\hat i+3\hat j-4\hat k)=0$
$\Rightarrow\:6(6\lambda+3)+3(3\lambda+1)-4(-4\lambda-10)=0$
$\Rightarrow\:\lambda=-1$
$\therefore Q$ is given by $(-4,0,0)$
$\therefore$ The required distance $AQ=\sqrt {9+4+36}=7$