Find k with the given vectors and if $\Vert ku + v\Vert = 13$

Question

Let $\mathbb{R}^3$ have the Euclidean inner product and $u=(1,1,-1)$ and $v=(6,7,-15)$.
Find k if $\Vert ku + v\Vert = 13$

So what I did was this:
$ \Vert ku + v\Vert^2 = \langle ku+v, ku + v\rangle = 13^2\\ \quad\quad\ \ 169=\langle ku,ku \rangle + 2\langle ku,v \rangle + \langle v,v \rangle \\ \quad\quad\ \ 169=k^2(1+1+1) + 2k(6+7+15) + (36+49+225) \\ \quad\quad\ \ 169=3k^2 + 56k + 310 \\ \quad\quad\ \ 0 = 3k^2 + 56k +141$

Then I just worked it out with a quadratic formula which came to. 15.49 or 3.17. I'm thinking this is not correct and there is no answer in the book. Does anybody have any ideas with this one?

Answer 1

$k (1,1,-1)+(6,7,-15)=(6 + k, 7 + k, -15 - k)$

$(6+k)^2+(7+k)^2+(-15-k)^2=169$

$3 k^2+56 k+141=0$

$k=-\dfrac{47}{3},\;k=-3$

Hope this helps