jmc

We expand $(ac+bd{)}^2+(bc-ad{)}^2$ :

$$\begin{array}{rl}(ac+bd{)}^2+(bc-ad{)}^2& =(ac{)}^2+2(ac)(bd)+(bd{)}^2+(bc{)}^2-2(bc)(ad)+(ad{)}^2\\ & =a^2{c}^2+2abcd+b^2{d}^2+b^2{c}^2-2abcd+a^2{d}^2\\ & =a^2{c}^2+b^2{d}^2+b^2{c}^2+a^2{d}^2\\ & =a^2{c}^2+b^2{c}^2+b^2{d}^2+a^2{d}^2.\end{array}$$Now we can factor out $c^2$ from the first two terms and $d^2$ from the last two terms and we get:

$$a^2{c}^2+b^2{c}^2+b^2{d}^2+a^2{d}^2=c^2(a^2+b^2)+d^2(b^2+a^2).$$Now we use distributive property and we have $$c^2(a^2+b^2)+d^2(b^2+a^2)=(c^2+d^2)(a^2+b^2)=4\cdot 3=\boxed{12}.$$