One way to approach this problem is with the theory of vectors.
Let $(a,b,c)$ repeesent an ordered triple where $a=x_1,b=x_2,c=x_3$. Define $n(a,b,c)=(na,nb,nc)$.
Clearly one solution is $(1,-1,1)$. Other solutions must then satisfy the superposition relation
$(x_1,x_2,x_3)=(1,-1,1)+(x_1^*,x_2^*,x_3^*)$
where the second term satisfies the homogeneous relation
$9x_1^*+12x_2^*+16x_3^*=0$
The solutions to the homogeneous equation form a two-dimensional vector space, they are orthogonal to $(9,12,16)$. So it is sufficient to find two linearly independent solutions. One is obtained by looking at the case $x_3^*=0$:
$9x_1^*+12x_2^*=0, (x_1^*,x_2^*,x_3^*)=n_1(4,-3,0)$
A second linearly independent soution is obtained from the case $x_1^*=0$:
$12x_2^*+16x_3^*=0, (x_1^*,x_2^*,x_3^*)=n_2(0,4,-3)$
Now superimpose the two independent homogeneous solutions with the partucular solution $(1,-1,1)$ and get:
$x_1=1+4n_1$
$x_2=-1-3n_1+4n_2$
$x_3=1-3n_2$.