$X_{1},X_{2},X_{3}\overset{i.i.d.}{\sim}N(0,1)$, find m.g.f. of $Y=X_{1}X_{2}+X_{1}X_{3}+X_{2}X_{3}$ I tried this
$X_{1}X_{2}+X_{1}X_{3}+X_{2}X_{3}=X_{1}(X_{2}+X_{3})+\frac{1}{4}(X_{2}+X_{3})^{2}-\frac{1}{4}(X_{2}-X_{3})^{2}$
$U=X_{2}+X_{3}\sim N(0,2)$
$\psi_{X_{1}(X_{2}+X_{3})}(t)=\psi_{X_{1}U}(t)=\frac{1}{\sqrt{2}\cdot 2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{x_{1}ut}e^{-\frac{1}{2}(x_{1}^{2}+\frac{u^{2}}{2})}\, dx_{1}\, du$
$=\frac{1}{\sqrt{2}\cdot 2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\frac{1}{4}(2(x_{1}-ut)^{2}-2u^{2}t^{2}+u^{2})}\, dx_{1}\, du$
$V=(x_{1}-ut)\qquad dv=dx_{1}\\
        =\frac{1}{\sqrt{2}\cdot 2\pi}\int_{-\infty}^{\infty}e^{-\frac{1}{4}(1-2t^{2})u^{2}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}v^{2}}\, dv\, du\\$
$=\frac{1}{\sqrt{\pi}\cdot 2}\int_{-\infty}^{\infty}e^{-\frac{1}{4}(1-2t^{2})u^{2}}\, du$
$ w=\sqrt{\frac{1-2t^{2}}{2}}u,\qquad \sqrt{\frac{2}{1-2t^{2}}}\, dw=du\\
    \psi_{XU}(t)=\frac{1}{\sqrt{1-2t^{2}}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}w^{2}}\, dw=\frac{1}{\sqrt{1-2t^{2}}}$
$Z=X_{2}-X_{3}\sim N(0,2)$
$\psi_{\frac{1}{4}(X_{1}+X_{2})^{2}}(t)=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}e^{\frac{1}{4}z^{2}t}e^{-\frac{1}{4}z^{2}}\, dz$
$=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{4}(z^{2}-z^{2}t)}\, dz=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{4}(1-t)z^{2}}\, dz\\$
$v=\sqrt{1-t}z,\quad \frac{1}{\sqrt{1-t}}\, dv=dz\\
        \psi_{\frac{1}{4}(X_{1}+X_{2})^{2}}(t)=\frac{1}{\sqrt{1-t}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{4\pi}}e^{-\frac{1}{4}z^{2}}\,dz=\frac{1}{\sqrt{1-t}}$
$\psi_{\frac{1}{4}(X_{1}+X_{2})^{2}-\frac{1}{4}(X_{2}-X_{3})^{2}}(t)=\frac{1}{\sqrt{1-t}}\cdot \frac{1}{\sqrt{1-(-t)}}=\frac{1}{\sqrt{(1-t)(1+t)}}=\frac{1}{\sqrt{1-t^{2}}}$
$\psi_{X_{1}(X_{2}+X_{3})+\frac{1}{4}(X_{2}+X_{3})^{2}-\frac{1}{4}(X_{2}-X_{3})^{2}}(t)=\frac{1}{\sqrt{1-2t^{2}}}\cdot \frac{1}{\sqrt{1-t^{2}}}=\frac{1}{\sqrt{1-2t^{2}}\sqrt{1-t^{2}}}$
But I have learned that this is incorrect. What mistakes did I make?

In your first line you write the expression as the sum of three quadratic forms. You then multiply the three mgfs. But, this is only possible if the variables (quadratic forms) are independent. Quadratic forms are independent if and only if the product of their matrices is the zero matrix. This is not true of the first and second quadratic forms. However, it is true of the first and third and it is true of the second and third. That is why you could multiply two of the mgfs and get a correct result for those.
