Construct an orthogonal matrix having $\frac{1}{2}(1,1,1,1)$ as its first row.
Let $\bf A$ be the required matrix.
Looking at the first row, I can just construct a Helmert matrix which is necessarily orthogonal.
$$\mathbf A=\begin{pmatrix}\frac{1}{\sqrt{4}}&\frac{1}{\sqrt{4}}&\frac{1}{\sqrt{4}}&\frac{1}{\sqrt{4}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}&0&0\\\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{6}}&\frac{-2}{\sqrt{6}}&0\\\frac{1}{\sqrt{12}}&\frac{1}{\sqrt{12}}&\frac{1}{\sqrt{12}}&\frac{-3}{\sqrt{12}}\end{pmatrix}$$
Is this a unique answer? Can the row vectors of $\bf A$ form different orthonormal bases of $\mathrm V_R(\mathbf A)$, the row space of $\bf A$, keeping the first row intact?
I have to find $3$ vectors which are orthogonal to the $1$st row and also orthogonal to each other. Just using the definition and taking the $2$nd, $3$rd and $4$th row of the required matrix as $(x_1,...,x_4), (y_1,...,y_4)$ and $(z_1...,z_4)$ respectively I get from $\bf AA'=I$,
$$\sum_{i=1}^4x_i=\sum_{i=1}^4y_i=\sum_{i=1}^4z_i=0$$
$$\sum_{i=1}^4x_i^2=\sum_{i=1}^4y_i^2=\sum_{i=1}^4z_i^2=1$$
$$\sum_{i=1}^4x_iy_i=\sum_{i=1}^4x_iz_i=\sum_{i=1}^4y_iz_i=0$$
However I don't think this is the right way to proceed, searching for solutions by trial and error.
I would like to know the right approach for this particular problem and also in general, the easiest way to construct an orthogonal matrix given any one of its rows/columns.
Let $B$ be any $3$-by-$3$ orthogonal matrix and $C$ be the block matrix $$C=\pmatrix{1&0\\0&B}.$$ Then $CA$ is an orthogonal matrix with the same first row as $A$; you get all such matrices that way, so there's a lot of them!
My favourite example is the scaled Hadamard matrix $$\frac12\pmatrix{1&1&1&1\\1&1&-1&-1\\1&-1&1&-1\\1&-1&-1&1}.$$