We assume that the following determinant is equal to zero:
$$\det A=\begin{vmatrix} -x&1&1&1\\ 1&-y&1&1\\ 1&1&-z&1\\ 1&1&1&-t\\ \end{vmatrix}=0$$
with $x$, $y$, $z$, $t$ positive integers. Then,
$$\det A=(xt-1)(yz-1)-(x+t+2)(y+z+2)$$
I have to determine positive integers $a$, $b$, $c$, $d$, of the system $Ar=0$, with $r=[a;b;c;d]$ with $a≠b≠c≠d $. The system has not the trivial solution, but what do I have to think?
Consider the system of linear equations determining $(a,b,c,d)$: $$ \begin{cases} \begin{align} -xa+b+c+d&=0\\ a-yb+c+d&=0\\ a+b-zc+d&=0\\ a+b+c-td&=0 \end{align} \end{cases}\tag{1} $$ Subtracting pairwise the equations one obtains: $$ a:b:c:d=\frac{1}{1+x}:\frac{1}{1+y}:\frac{1}{1+z}:\frac{1}{1+t}.\tag{2} $$ Multiplying the RHS with $(1+x)(1+y)(1+z)(1+t)$ results in the integer solution: $$ \begin{cases} a=(1+y)(1+z)(1+t),\\ b=(1+x)(1+z)(1+t),\\ c=(1+x)(1+y)(1+t),\\ d=(1+x)(1+y)(1+z). \end{cases}\tag{3} $$ One can easily check that substitution of the values from (3) into any of equations (1) results in the expression for $\det A$, i.e. $0$.
As seen from (2) the condition $a\ne b\ne c\ne d$ implies $x\ne y\ne z\ne t$. Though it is not immediately obvious, by brute force one finds that six such solutions indeed exist: $$ (1,2,6,41),\\ (1,2,7,23),\\ (1,2,8,17),\\ (1,2,9,14),\\ (1,3,4,19),\\ (1,3,5,11). $$ The corresponding vectors $(a,b,c,d)$ can be computed from (3). For example for $(1,3,5,11)$ it is: $$ (4\times6\times12,\,2\times6\times12,\,2\times4\times12,\,2\times4\times6)\sim(6,3,2,1). $$