Given that, $a_1=2 \hat{i}-\hat{j}+\hat{k}$
$\begin{aligned}
& a_2=3 \hat{i}-4 \hat{j}-4 \hat{k} \\
& a_3=-\hat{i}+\hat{j}-\hat{k} \\
& a_4=-\hat{i}+3 \hat{j}+\hat{k}
\end{aligned}$
As $M_1, M_2, M_3$ and $M_4$ are the magnitudes of the vectors $\mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3$ and $\mathrm{a}_4$ respectively.
$\begin{aligned}
& \therefore \quad M_1=\left|\mathrm{a}_1\right| \\
& =\sqrt{(2)^2+(-1)^2+(1)^2} \\
& =\sqrt{4+1+1}=\sqrt{6} \\
& M_2=\left|\mathrm{a}_2\right| \\
& =\sqrt{(3)^2+(-4)^2+(-4)^2} \\
& =\sqrt{9+16+16}=\sqrt{41} \\
& M_3=\left|\mathrm{a}_3\right| \\
& =\sqrt{(-1)^2+(1)^2+(-1)^2} \\
& =\sqrt{1+1+1}=\sqrt{3} \\
& M_4=\left|\mathrm{a}_4\right|
\end{aligned}$
$\begin{aligned}
& =\sqrt{(-1)^2+(3)^2+(-1)^2} \\
& =\sqrt{1+9+1}=\sqrt{11}
\end{aligned}$
Hence, the correct order of magnitudes
$M_3 < M_1 < M_4 < M_2$