Vector of Steepest Ascent

First, the orthogonality condition requires that

$\left(1,3,2\right) \cdot \left(a,b,c\right) = a+3b+2c=0$

This leads to the equation of the line where plane of possible vectors intersects with the plane $c=0$

$a+3b=0$

and the equation of the line perpendicular to this line is

$b=3a$

so, the vector we seek is

$\left( \dfrac{a}{k}, \dfrac{b}{k}, \dfrac{c}{k} \right)$

where $b=3a$
$c=\dfrac{1}{2}\left(-a-3b\right)$
and $k=\sqrt{ {a}^{2}+{b}^{2}+{c}^{2} }$

Letting $a=1$ , we have the vector

$\left( -\dfrac{1}{\sqrt{35}}, -\dfrac{3}{\sqrt{35}}, \sqrt{ \dfrac{5}{7} } \right)$

and so the answer works out to $169$

Imagine a plane through the origin, normal to the given vector (which is in the first octant). Imagine a vertical plane also through the origin. The vector we seek will be along the intersection of these planes and will be in the third octant.

Imagine their projections on the xy plane. They will be equal and opposite: <-1,-3,z>

Equate the dot product to zero: <1,3,2>.<-1,-3,z> = 0 = -1-9-2z. Giving z = 5. Normalizing <-1,-3,5> gives the desired unit vector $\frac{<-1,-3,5>}{\sqrt{35}}$

And the desired quantity = $\frac{1000(-1-3+5)}{\sqrt{35}}= \frac{1000}{\sqrt{35}} = 169.031$