By definition, an isometric matrix has the property $A^*A = AA^* = 1$, with $A^*$ as the hermitian transpose of $A$. So we construct the matrix $$A = \begin{bmatrix} 1 & 1 & 1 \\ a & b & c\\ d & e & f\end{bmatrix}$$ and have $A^*A = 1$, which gives us $$\begin{bmatrix}3&a + b + c&d + e + f\\ a + b + c&a^2 + b^2 + c^2&ad + be + cf \\ d + e + f&ad + be + cf&d^2 + e^2 + f^2 \end{bmatrix} = 1 \tag{1}$$.
So first of all I have no idea how to solve this...
Second of all, as a technicality, I assume $1$ in the equation $A^*A = AA^* = 1$ is the identity matrix, so how can the matrix in $(1)$ be the identity matrix if its first entry is $3$? If all of the elements in the diagonal are 3, then this would be $3I$. Which is still not $1$.
Thanks a bunch.
As a row (or a column) of any isometry matrix (aka "unitary" or orthogonal) has to be unit norm, $(1,1,1)$ cannot work.
We have to find a first row with the form $(k,k,k)$ such that its (squared) norm is $1$, giving condition $3k^2=1$, thus $k=\pm 1/\sqrt{3}$.
Let us take $k=1/\sqrt{3}$ and look for matrices with this fixed first row:
$$A = \begin{bmatrix} 1/\sqrt{3} & 1/\sqrt{3} & 1/\sqrt{3} \\ a & b & c\\ d & e & f\end{bmatrix}$$
There is a large choice of a second row giving a zero dot product with the first, moreover with unit norm, for example by taking :
$$A = \begin{bmatrix} 1/\sqrt{3} & 1/\sqrt{3} & 1/\sqrt{3} \\ 1/\sqrt{2} & -1/\sqrt{2} & 0\\ d & e& f\end{bmatrix}$$
Now, the third row. It must
(a) be orthogonal to the first and the second row, and moreover
(b) be unit norm.
Condition (b) is not difficult to fulfill (one divides by the norm : in fact, this will not be necessary).
Concerning condition (a), it suffices to take the cross product of the 2 first rows...
$$\begin{bmatrix} d\\e\\f\end{bmatrix} \ = \ \begin{bmatrix} 1/\sqrt{3}\\1/\sqrt{3}\\1/\sqrt{3}\end{bmatrix} \ \times \ \begin{bmatrix} 1/\sqrt{2}\\-1/\sqrt{2}\\0\end{bmatrix}=\begin{bmatrix} 1/\sqrt{6}\\1/\sqrt{6}\\- 2/\sqrt{6}\end{bmatrix}$$
Thus an answer is :