You did not describe what the "original" orientation of the camera
actually is, that is, what the unrotated camera looks like.
I'll suppose that the camera would be pointing along the positive $x$-axis
and that a vector from the center to the "top" of the camera would point
in the direction of the positive $z$-axis.
Now let's say you want the camera pointing downward at a $45$-degree angle
in the $x,z$ plane, that is, the axis of the camera will be
in the direction of the vector
$\left[\frac{\sqrt2}{2}, 0, -\frac{\sqrt2}{2}\right]$,
and that the direction to the "top" of the camera will
also still be in the $x,z$, plane, in the direction
$\left[\frac{\sqrt2}{2}, 0, \frac{\sqrt2}{2}\right]$.
You get the camera into this orientation by rotating it $45$ degrees
($\pi/4$ radians)
around the $y$-axis, that is, using the rotation quaternion
$\left[0, \sin\left(\frac\pi8\right), 0, \cos\left(\frac\pi8\right)\right]$
(or maybe that should be $-\sin\left(\frac\pi8\right)$ instead of
$\sin\left(\frac\pi8\right)$; it depends on which way is a
"positive" rotation around the axis).
In short, the axis of rotation that turns your camera to the desired orientation
is usually nowhere near the axis along which the camera ends up pointing.
The axis and angle to achieve a general orientation is not usually as
obvious as in this simple case. Rather than trying to guess it directly,
you are probably better off performing a series of simple rotations,
for example elevate or depress the angle of the camera and then rotate
it around the vertical axis (the $z$ axis in my example, or whichever of
the axes is vertical in your system). If you also want the camera
to be "twisted" around its own axis, then you might want to
rotate it about that axis before the other two rotations.
To represent a sequence of rotations,
you multiply together the quaternions for each rotation.
You then have a single quaternion that represents a single rotation
taking the camera from its original orientation to the desired orientation.
I can answer this:
What is the affine transformation converting world coordinates to camera coordinates? (camera world coordinates: $c=(c_x,c_y,c_z)^\top$, visual center world coordinates, $v=(v_x,v_y,v_z)^\top$)
I'm assuming the traditional camera image coordinates (before projection) having $z$ drilling "into" the image, $x$ pointing from left to right, and $y$ pointing downward.
Now let's track how the axes must be rotated without translation:
1. the new $z$ axis ($z'$) will point along $v-c$.
1. the new $x$ axis ($x'$) is perpendicular to $z$ and $z'$
1. the new $y$ axis ($y'$) is perpendicular to $x'$ and $z'$.
You can find three vectors that point along the new axes in world coordinates, normalize them, then put them in the rows of a $3\times 3$ matrix $R$: this converts world coordinates to rotated camera orientation.
Finally, if you know the translation $t$ in world coordinates (it would be $(-10,-10,-10)^\top$ to translate to the camera's position in world coordinates) then the translation in camera coordinates is $t'=Rt$
Let's actually carry this out for your example. Let's work on a triad of orthogonal vectors:
$z'=(-1,-1,-1)$, pointing in the direction the camera must face.
$x'=z'\times z=(-1,1,0)^\top$
$y'=z'\times x'=(1,1,-2)^\top$
Normalizing these and using them as the rows of a matrix you get:
$$
R=\frac{1}{\sqrt{6}}\begin{bmatrix}
-\sqrt{3}&\sqrt{3}&0\\
1&1&-2\\
-\sqrt{2}&-\sqrt{2}&-\sqrt{2}
\end{bmatrix}
$$
Then $t'=Rt=(0,0,10\sqrt{3})$.
Notice that the angle of declination is an odd angle near $35^\circ$ rather than exactly $45^\circ$. (I had a hard time seeing this at first, but if you draw a cube and check the angle between $(1,1,0)$ and $(1,1,1)$ you'll see what I mean.)
Now you've converted world coordinates to rotated frame that is aligned with your camera's frame, but differs by a translation.
This gives you the resulting affine transformation $\begin{bmatrix}R&t'\\0_{1\times 3}&1\end{bmatrix}$
which carries world coordinates to camera coordinates.
As a sanity check, you can confirm that the world's origin maps to camera $(0,0,10\sqrt{3})^\top$ and that the world camera location $(10,10,10)$ now maps to the camera's origin. A third check of your choice should be sufficient to convince you this is the right $R$ and $t'$.
One caveat: I'm not 100% sure the step with $z\times z'$ is always in this order. I picked it this way on this occasion because it gave the right orientation for $x'$ and $y'$ in the end. Hopefully that is all consistent, but maybe there is some sign ambiguity after all.
The second question is how to construct the "UP" vector.
I don't understand what you are asking. If you mean the camera coordinates for the direction of the world $z$-axis, then that would just be $R(0,0,1)^\top +t'$.
Finally, I will have to rotate camera as well from "landscape" to "portrait" orientation .
I'm interpreting this to mean that you'd want to rotate the image plane so that the $y$-axis is horizontal, which could be done with a $\pi/4$ rotation in either way around the camera $z$-axis.
This transformation should be entirely obvious:
$$U=
\begin{bmatrix}
0&-1&0\\
1&0&0\\
0&0&1\end{bmatrix}
$$
$U$ gives the rotation in the clockwise direction around the $z$ axis (which would look to be counterclockwise if you are looking up the $z$ axis into the picture) and $U^\top$ would give the rotation in the other direction.
Best Answer
If Euler vector is the same as Euler angles. Than is is not difficult to construct transformation from local space of plane to world space.
$v_{world} = R v_{local} + p_{plane}$
Where $R$ is rotation matrix qiven by Euler angles.
see:http://en.wikipedia.org/wiki/Rotation_formalisms_in_three_dimensions#Conversion_formulae_between_formalisms
$p_{plane}$ is position of plane.
$v_{local}$ is given position of camera.
$(\phi,\theta,\psi)$ is your Euler vector. Than $R = A_3 A_2 A_1 $ (matrices $A_i$ are those from wiki link)