There is a $3\times 3 $ orthogonal matrix with all non zero entries.? Is the statement true?
There is a $3\times 3 $ real orthogonal matrix with all non zero entries.
for orthogonality, $AA^T=A^TA=I_3$, please give me hint

Gordon Pall found all rational orthogonal matrices, 3 by 3, in 1940, see PALL_PDF
Given an odd number
$$ n = a^2 + b^2 + c^2 + d^2,$$ the matrix
$$ \frac{1}{n} \begin{pmatrix} a^2 + b^2 - c^2 - d^2 & 2(-ad+bc) & 2(ac+bd) \\ 2(ad+bc) & a^2 - b^2 + c^2 - d^2 & 2(-ab+cd) \\ 2(-ac+bd) & 2(ab+cd) & a^2 - b^2 - c^2 + d^2 \end{pmatrix}$$
is orthogonal, and all rational orthogonal matrices can be written in this manner. This is formula (10) on page 755 of Pall's article,  the second page of the pdf.
Right, so you want no zeroes in the matrix. Take, for example, $d=0,$  and nonzero $a,b,c$ which do not make a Pythagorean triple in any order. It would be enough to just add on the condition $c^2 \geq 1 + a^2 + b^2.$
