Given curve $x^2+y^2-2 x-4 y+2=0$ and line $x+y=k$
$$
\begin{gathered}
x+y=k \\
\frac{x+y}{k}=1
\end{gathered}
$$
Now, Homogenising (i) and the given curve.
Take, $x^2+y^2-2(x+2 y)(1)+2(1)^2=0$
$$
\begin{aligned}
& x^2+y^2-2(x+2 y) \frac{(x+y)}{k}+2\left(\frac{x+y}{k}\right)^2=0 \\
& x^2+y^2-\frac{2\left(x^2+3 x y+2 y^2\right)}{k}+\frac{2\left(x^2+y^2+2 x y\right)}{k^2}=0 \\
& k^2 x^2+k^2 y^2-2 k x^2-6 k x y-4 k y^2+2 x^2+2 y^2+4 x y=0 \\
& x^2\left(k^2-2 k+2\right)+y^2\left(k^2-4 k+2\right)+x y(-6 k+4)=0
\end{aligned}
$$
When lines are at right angles then sum of coefficients of
$$
\begin{aligned}
& x^2 \& y^2 \text { is } 0 . \\
& \text { So, } k^2-2 k+2+k^2-4 k+2=0 \\
& 2 k^2-6 k+4=0 \\
& k^2-3 k+2=0 \\
& k^2-2 k-k+2=0 \\
& (k-2)(k-1)=0 \\
& k=2,1
\end{aligned}
$$
Possible sum of values of $k=1+2=3$.
So, option (c) is correct.