Answer
See answer below
Work Step by Step
We are given $p=2-x-x^2\\
q=1+x+x^2\\
r=a+bx+cx^2$
$p,q$ and $r$ are orthogonal if
$(p,q)=2.1+(-1).1+(-1).1
=0$
$(p,r)=(2-x-x^2,a+bx+cx^2)\\
=2.a+(-1).b+(-1).c\\
=2a-b-c\\
=0$
$(q,r)=(1+x+x^2,a+bx+cx^2)\\
=1.a+1.b+1.c\\
=a+b+c\\
=0$
We have a system of equations:
$2a-b-c=0\\
a+b+c=0$
Adding first equation to the second equation we have:
$3a=0 \\
a=0\\
\rightarrow b+c=0 \rightarrow c=-b$
Hence, the set $\{p(x),q(x),r(x)$ is an orthogonal set if $r(x)=bx-bx^2$ with $b \in R$
