Using Lagrange multipliers to maximize directional derivative 
I've tried to use Lagrange's multipliers to solve this question, but I'm unable to find an answer that matches the choices. I know that the gradient, i.e. $\triangledown f \langle x, y, z \rangle$ would be $\langle -1, 2, 3 \rangle$, and that the equation of the tangent plane would be $x + z -2 =0$, but I don't know how to use the constraint. I know that the unit vector that I multiply the gradient with must lie on this plane, but what does that mean quantitatively? 
I've taken the unit vector to be $\langle a, b, c\rangle$, so that tells me that $-a + 2b + 3c = 0$.
I also know that since $\langle a, b, c\rangle$ lies on the plane $x+z=2$, this means that $a+c=2$, and since it is a unit vector, $| \langle a, b, c\rangle| = 1$, which means that $a^2 + b^2 + c^2 =1$. I've tried to solve the double constraint, but my answer does not match the answer choices. 
 A: The normal vector to the tangent plane $T$ is $\parallel OP$, i.e., ${\bf n}=(1,0,1)$. We have to find two unit vectors in $T$ that are orthogonal ${\bf n}$ and to each other. These are for instance the vectors
$${\bf a}=(0,1,0),  \qquad {\bf b}=\left({1\over\sqrt{2}},0,-{1\over\sqrt{2}}\right)$$
(found by guessing), so that the general admissible vector ${\bf u}\in T$ is given by
$${\bf u}=\cos\phi\> {\bf a}+\sin\phi\> {\bf b}\ .$$
Now
$$D_{\bf u}f(P)=\langle \nabla f(P),{\bf u}\rangle=\left\langle(-1,2,3),\left({\sin\phi\over\sqrt{2}},\cos\phi,-{\sin\phi\over\sqrt{2}}\right)\right\rangle\ .$$ 
It follows that
$$D_{\bf u}f(P)=2\cos\phi-2\sqrt{2}\>\sin\phi\ ,$$
so that the maximum is
$$\sqrt{2^2+\bigl(2\sqrt{2}\bigr)^2}=2\sqrt{3}\ .$$
If you want to use Lagrange multipliers then you have to find a vector ${\bf u}=(\xi,\eta,\zeta)$ satisfying
$$\langle{\bf u},{\bf n}\rangle=\xi+\zeta=0,\qquad\xi^2+\eta^2+\zeta^2=1$$
such that 
$$\langle\nabla f(P),{\bf u}\rangle=-\xi+2\eta+3\zeta$$
is maximal. This will lead to a lot of equations.
