Answer
$\frac{4\sqrt 4929}{53}$
Work Step by Step
We need to form vector $v$ passes through $(0,0,0),(6,-1,-4)$ and $w$ from $(0,0,0)$ to $(2,3,4)$ first:
$v=(6-0,-1-0,-4-0)=(6,-1,-4) \\
w=(2-0,3-0,4-0)=(2,3,4)$
The distance from $(2,3,4)$ to the line in $R^3$ is:
$$||w-P(w,v)|| $$
Thus:
$$w-P(w,v)=w-\frac{(w,v)}{||v||^2}v\\
=(2,3,4)-\frac{}{||(6,-1,-4)||^2}(6,-1,-4) \\
=(2,3,4)-\frac{2.6+3.(-1)+4.(-4)}{6^2+(-1)^2+(-4)^2}(6,-1,-4)\\
= (2,3,4) -\frac{7}{33}(6,-1,-4)\\
=\frac{4}{53}(37,38,46)$$
Hence, we have: $d=||w-P(w,v)||=\sqrt (\frac{4}{53})^2(37^2+38^2+46^2)=\frac{4\sqrt 4929}{53}$
