Let $f:\mathbb{R}^3\rightarrow \mathbb{R}$ be the function defined as $f(x,y,z)=ax^2y+by^2 z+cz^2 x$. Find the constants $a,b,c$ such that in the point $(1,1,1)$, the directional derivative is maximum in the direction $\hat{u}=\frac{1}{\sqrt{26}}(1,5,0)$ and equals $13$.
So I know that the gradient can be found $\vec{\nabla f (1,1,1)}=(2a+c,2b+a,2c+b)$, and then I know since it is a differentiable function that: $$f'_{u}(1,1,1)=13=\vec{\nabla f(1,1,1)}\cdot \hat{u}\Rightarrow 13=(2a+c,2b+a,2c+b)\cdot (1,5,0)\frac{1}{\sqrt{26}}$$ but then I have an equation with 3 variables and I don't know how to continue. Can anyone shine some light on this?