$\underline{\textbf{Answer:}\Rightarrow}\;1)\;160$
$\underline{\textbf{Explanation:}\Rightarrow}$
$\textbf{Given:}\;7^{17}\;\text{mod}\;561$
$\mathbf{561}$ can be factorised into $3\times 17\times 11$.
Now,
$7^{17} = 3\times \mathbf{q_1}+1\\7^{17} = 17\times \mathbf{q_2}+7\\7^{17}= 11\times \mathbf{q_3} - 5\;\;\;\text{[$\therefore\;$Positive remainder = 6]}$
Now choose among the all options which when divided by $\mathbf{3,17,11}$ gives the remainder $\mathbf{1,7,6}$.
$\therefore$ Only option $(1)$ satisfies.
$\textbf{Calculating remainders using Euler's Theorem:}$
$\mathrm{\begin{align} 7^{17} \;\text{mod}\;11 &= 7^{10}\times 7^7 \;\text{mod} \;11 \\&= 1 \times 7^7 \;\;\text{mod} \;\;11 \\&= 7^3 \times 7^3 \times 7\;\;\text{mod} \;\;11\\&= 343\;\;\text{mod}\;\;11\times 343\;\;\text{mod}\;\;11\times 7\;\;\text{mod}\;\;11 \\&= 2\times 2\times7\;\;\text{mod} \;\;11 \\&= 28\;\;\text{mod}\;\;11 \\&= 6\;\;\text{mod} \;\;11 \\&= \color{magenta}{\enclose{circle}{-5}} \; \;\text{mod} \;\;11 \end{align}}$
Or, remainder $=\color{magenta}{\enclose{circle}{6}}\;\;\;\mathrm{[\because 6+5=11]}$
Similarly, the other two remainders can be calculated.
$\therefore\;\textbf{1)}$ is the right option.
$\textbf{For extended Reading, Refer:}$
http://mathonline.wikidot.com/examples-of-finding-remainders-of-large-numbers