Compute the volume of the parallelepiped on vec m=vec a+vec…
Compute the volume of the parallelepiped on $\vec m=\vec a+\vec b+\vec c$, $\vec n=\vec a+\vec b-\vec c$, $\vec p=\vec a-\vec b+\vec c$, if $\vec a,\vec b,\vec c$ form a right-handed triad and the parallelepiped on them has volume $\tfrac14$.
Explanation of Method
Volume of the parallelepiped on m, n, p expressed through a, b, c.
Step 1: Coefficient determinant
The triple product is linear in each slot. Every term with a repeated vector vanishes, and the six remaining terms combine exactly like a $3\times3$ determinant. Hence
\[ [\vec m,\vec n,\vec p]=\begin{vmatrix}1&1&1\\1&1&-1\\1&-1&1\end{vmatrix}[\vec a,\vec b,\vec c]=\big(1(1-1)-1(1+1)+1(-1-1)\big)[\vec a,\vec b,\vec c]=-4\,[\vec a,\vec b,\vec c]. \]
Step 2: Volume
The triad is right-handed, so $[\vec a,\vec b,\vec c]=+\tfrac14$, and
\[V'=|-4\cdot\tfrac14|=1\]
The negative sign means that $\vec m,\vec n,\vec p$ (in this order) form a left-handed triad.
Result
\[\boxed{\,V'=|-4\cdot\tfrac14|=1\,}\]
Remark
Cleaned up. The source's permutation table contains contradictory rows and a "re-doing" pass. The determinant rule above gives the result directly, and it matches the source's own verification ($-4$).