Let us assume, the length of spring be l.

When we cut the spring into ratio of length 1 : 2 : 3, we get three springs of lengths ${l \over 6},{{2l} \over 6}$ and ${{3l} \over 6}$ with force constant,

$\therefore$ ${k_1} = {{kl} \over {{l_1}}} = {{kl} \over {l/6}} = 6k$

${k_2} = {{kl} \over {{l_2}}} = {{kl} \over {2l/6}} = 3k$

${k_3} = {{kl} \over {{l_3}}} = {{kl} \over {3l/6}} = 2k$

When connected in series,

${1 \over {k'}} = {1 \over {6k}} + {1 \over {3k}} + {1 \over {2k}} = {{1 + 2 + 3} \over {6k}} = {1 \over k}$

$\therefore k' = k$

When connected in parallel,

k'' = 6k + 3k + 2k = 11k

${{k'} \over {k''}} = {k \over {11k}} = {1 \over {11}}$