12. Given four integers $a, b, c, d$ are all even, and $0<a<b<c<d, d-a=90$, if $a, b, c$ form an arithmetic sequence, and $b, c, d$ form a geometric sequence, then $a+b+c+d=$ $\qquad$ .
Let the four numbers be $b-m, b, b+m, \frac{(b+m)^{2}}{b}$, then $\frac{(b+m)^{2}}{b}-b+m=90$ $\Rightarrow m(m+3 b)=90 b$. Since $b, m$ are both even, $m+3 b \equiv m(\bmod 6)$, thus $m, m+3 b$ are both multiples of 6. Also, $0<m<b \Rightarrow 3 b<m+3 b<4 b \Rightarrow \frac{90}{4}<m<30 \Rightarrow m=24$, hence $24(24+3 b)=90 b \Rightarrow 8+b=\frac{5}{4} b \Rightarrow b=32$.

Therefore, $a+b+c+d=8+32+56+98=194$.