= Solution
An <orthogonal matrix> satisfies $R^TR=I$, so it preserves the <Euclidean norm>. For a nonzero real <eigenvector> $v$,
$$
\|v\|^2=\|Rv\|^2=\|\lambda v\|^2=\lambda^2\|v\|^2.
$$
Therefore $\boxed{\lambda=\pm1}$. Geometrically, <eigenvalue> $1$ fixes every vector on the <linear span> of $v$; <eigenvalue> $-1$ reverses every vector on that line. In either case the line is preserved as a set.
For $M$, solving $(M-I)v=0$ gives $y=x$, $z=y$, $x=z$. For $N$, the equations are $-2y-2z=0$ and $-2z=0$, hence $y=z=0$. For $P$, let $a=(1,1,1)^T$ and $J=aa^T$, so that $P=I-\frac23J$. Then $(P-I)v=0$ is equivalent to $x+y+z=0$. The resulting <eigenspaces> and their <bases> are
$$
\boxed{\begin{aligned}
E_1(M)&=\operatorname{span}\{(1,1,1)^T\},&\dim E_1(M)&=1,\\
E_1(N)&=\operatorname{span}\{(1,0,0)^T\},&\dim E_1(N)&=1,\\
E_1(P)&=\operatorname{span}\{(1,-1,0)^T,(1,0,-1)^T\},&\dim E_1(P)&=2.
\end{aligned}}
$$
Solving all the equations shows that these lists contain the maximum possible numbers of independent <eigenvectors> with <eigenvalue> $1$.
The columns of $M$ form an <orthonormal basis>, so $M$ is an <orthogonal matrix>. Also $\det M=1$. Its fixed axis is the line through $a$, and on the perpendicular plane it acts as a two-dimensional rotation. If $\theta$ denotes the rotation angle, its <matrix trace> is $1+2\cos\theta$. Since $\operatorname{tr}M=0$, we obtain $\cos\theta=-1/2$. To fix the orientation relative to $u=a/\sqrt3$, note that $Me_1=e_3$ and
$$
u\cdot(e_1\times Me_1)=u\cdot(e_1\times e_3)=-\frac1{\sqrt3}<0.
$$
The component of $e_1$ perpendicular to $u$ has squared length $2/3$, and this <scalar triple product> equals $(2/3)\sin\theta$. Hence $\sin\theta=-\sqrt3/2$. Thus
$$
\boxed{M:\ \text{axis }\mathbb R(1,1,1),\quad
\theta=-\frac{2\pi}{3}\text{ about }\frac{(1,1,1)}{\sqrt3}.}
$$
Equivalently, its unsigned angle is $2\pi/3$, or its positive angle is $2\pi/3$ about the opposite oriented axis.
For $P$, the <reflection in a hyperplane> formula gives
$$
P=I-2\frac{aa^T}{a^Ta}.
$$
It fixes the plane $a\cdot v=0$ pointwise and maps $a$ to $-a$, so $\boxed{P\text{ reflects in }x+y+z=0}$. These actions also show that $P$ is an <orthogonal matrix>. In contrast, the second column of $N$ has squared length $5$, so $N$ is not <orthogonal> and cannot represent either of these Euclidean motions.
A <diagonalizable matrix> must have a <basis> of <eigenvectors> over the field in question. For $M$, a direct <determinant> computation gives $\chi_M(t)=1-t^3$. Its <eigenvalues> are $1,\omega,\omega^2$, where $\omega=e^{2\pi i/3}$. Only one is real, with one-dimensional <eigenspace>, so $M$ is not diagonalizable over $\mathbb R$. Over $\mathbb C$ the three <eigenvalues> are distinct, and the vectors $(1,\lambda,\lambda^2)^T$ for $\lambda=1,\omega,\omega^2$ give an <eigenbasis>. The upper triangular <matrix> $N$ has only <eigenvalue> $1$, of <algebraic multiplicity> $3$, but its <eigenspace> has dimension $1$ over either field. Hence $N$ is not diagonalizable over either field. For $P$, the two real plane vectors displayed above, together with $a$, form an <eigenbasis>, with <eigenvalues> $1,1,-1$. They remain independent over $\mathbb C$. Therefore
$$
\boxed{\begin{array}{c|cc}
&\mathbb R&\mathbb C\\\hline
M&\text{not diagonalizable}&\text{diagonalizable}\\
N&\text{not diagonalizable}&\text{not diagonalizable}\\
P&\text{diagonalizable}&\text{diagonalizable}
\end{array}}
$$
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