Solution
= Solution
{c}
The identity $Y=aX+W$ forces $b=1-2a$ and $c=1-a$. Independence will follow from zero <covariance>, because $X,W$ are jointly <Gaussian random variables>. That covariance is
$$
\operatorname{Cov}(X,W)=2b+c=3-5a.
$$
Thus
$$
\boxed{a=\frac35,\qquad b=-\frac15,\qquad c=\frac25.}
$$
These values give $W=(-Z_1+2Z_2)/5$ with variance $1/5$ and zero covariance with $X$. Joint normality then proves the requested independence.