Given vector
\(\begin{aligned}
& \mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\
& \mathbf{b}=\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \text { and } \mathbf{c}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}
\end{aligned}\)
and as we know that
\(\left[\begin{array}{lll}\mathbf{a} \times \mathbf{b} & \mathbf{b} \times \mathbf{c} & \mathbf{c} \times \mathbf{a}\end{array}\right]=\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}\right]^2\)
and \(\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}\right]=\left|\begin{array}{ccc}2 & -3 & -4 \\ 1 & 4 & -2 \\ 3 & -1 & 4\end{array}\right|\)
\(\begin{aligned}
& =2(16-2)+3(4+6)-4(-1-12) \\
& =28+30+52=110
\end{aligned}\)
\(\therefore\left[\begin{array}{lll}\mathbf{a} \times \mathbf{b} & \mathbf{b} \times \mathbf{c} & \mathbf{c} \times \mathbf{a}\end{array}\right]=\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}\right]^2=(110)^2=12100\)
Hence, option (d) is correct.