It is given that,
$$
\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}
$$
and $\mathbf{c}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
So, vectors
$$
\begin{aligned}
\mathbf{a}+\mathbf{b}+\mathbf{c} & =6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\
\mathbf{a}-\mathbf{b}+\mathbf{c} & =4 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+7 \hat{\mathbf{k}} \\
\text { and } \quad \mathbf{a}+\mathbf{b}-\mathbf{c} & =0 \hat{\mathbf{i}}-0 \hat{\mathbf{j}}+\hat{\mathbf{k}}
\end{aligned}
$$

So, the required volume of the parallelopiped having $\mathbf{a}+\mathbf{b}+\mathbf{c}, \mathbf{a}-\mathbf{b}+\mathbf{c}$ and $\mathbf{a}+\mathbf{b}-\mathbf{c}$ as coterminus edges is
$$
v=\left\|\begin{array}{ccc}
6 & -2 & 3 \\
4 & -6 & 7 \\
0 & 0 & 1
\end{array}\right\|=|1(-36+8)|=|-28|=28 \text {. }
$$

Hence, option (c) is correct.