Question: A 3D line is given by 

$$r(t) = (8, 8, 9) + t(9, 4, 5), t ∈ R $$

Let $P(x, y, z)$ be the point on the line that is nearest to the the origin. Find the point $(x, y, z)$.

I know I can find the normal, **n** to vector **v**(9,4,5) by doing cross product of **u** x **v** where **u** is just origin to point (8,8,9). 
By doing cross product of **u** x **v**, I got my **n** to be (-5,41,-40). From here, I compare the ||**n**|| which is the vertical distance by finding another normal to the line which is OP x **v**. 

If I equate this ||**n**|| = ||OP x **v**||, I would get 

$$||(9y-8z, 9x+8z ,8x-8y)|| = ||(-5,41,-40)||$$

Finally, by simultaneous equation where by 

$$9y-8z=-5,\>\>\> 9x+8z=41,\>\>\>8x-8y=-40$$

and solve it to be $y=5,\>\>\> x=-1,\>\>\> z= 25/4$

which is the wrong answer. The answer to the question is 

$$x=-365/122,\>\>\> y= 190/61, \>\>\>z = 353/122$$

But,I simply can't figure this out.

Attached: ([A sketch on this question](https://i.stack.imgur.com/tEEGp.png)) which I'm not sure is accurate or not.