Consider the cube in the first octant with sides OP,OQ and OR of length 1, along the x-axis, y-axis and z-axis, respectively, where `O(0,0,0)` is the origin. Let `S(1/2,1/2,1/2)` be the centre of the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal OT. If `vec(p)=vec(SP), vec(q)=vec(SQ), vec(r)=vec(SR)` and `vec(t)=vec(ST)` then the value of `|(vec(p)xxvec(q))xx(vec(r)xx(vect))| is `

Answer

To solve the problem, we need to find the value of \( |(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t})| \) where \( \vec{p} = \vec{SP} \), \( \vec{q} = \vec{SQ} \), \( \vec{r} = \vec{SR} \), and \( \vec{t} = \vec{ST} \).

### Step-by-Step Solution:

1. **Identify the Coordinates of Points:**
- The origin \( O \) is at \( (0, 0, 0) \).
- The center of the cube \( S \) is at \( \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) \).
- The vertex opposite to \( O \) is \( T \) at \( (1, 1, 1) \).
- The vertices \( P \), \( Q \), and \( R \) are at \( (1, 0, 0) \), \( (0, 1, 0) \), and \( (0, 0, 1) \) respectively.

2. **Calculate Vectors:**
- \( \vec{SP} = \vec{P} - \vec{S} = (1, 0, 0) - \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) = \left(\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}\right) \)
- \( \vec{SQ} = \vec{Q} - \vec{S} = (0, 1, 0) - \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) = \left(-\frac{1}{2}, \frac{1}{2}, -\frac{1}{2}\right) \)
- \( \vec{SR} = \vec{R} - \vec{S} = (0, 0, 1) - \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) = \left(-\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}\right) \)
- \( \vec{ST} = \vec{T} - \vec{S} = (1, 1, 1) - \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) = \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) \)

3. **Cross Products:**
- First, calculate \( \vec{p} \times \vec{q} \):
\[ \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \end{vmatrix} \]
- Expanding this determinant gives:
\[ = \hat{i} \left(-\frac{1}{2} \cdot -\frac{1}{2} - (-\frac{1}{2} \cdot \frac{1}{2})\right) - \hat{j} \left(\frac{1}{2} \cdot -\frac{1}{2} - (-\frac{1}{2} \cdot -\frac{1}{2})\right) + \hat{k} \left(\frac{1}{2} \cdot \frac{1}{2} - (-\frac{1}{2} \cdot -\frac{1}{2})\right) \]
- This simplifies to:
\[ = \hat{i} \left(\frac{1}{4} + \frac{1}{4}\right) - \hat{j} \left(-\frac{1}{4} - \frac{1}{4}\right) + \hat{k} \left(\frac{1}{4} - \frac{1}{4}\right) = \hat{i} \frac{1}{2} + \hat{j} \frac{1}{2} + 0\hat{k} \]

- Next, calculate \( \vec{r} \times \vec{t} \):
\[ \vec{r} \times \vec{t} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{vmatrix} \]
- Expanding this determinant gives:
\[ = \hat{i} \left(-\frac{1}{2} \cdot \frac{1}{2} - (-\frac{1}{2} \cdot \frac{1}{2})\right) - \hat{j} \left(-\frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{2}\right) + \hat{k} \left(-\frac{1}{2} \cdot \frac{1}{2} - (-\frac{1}{2} \cdot -\frac{1}{2})\right) \]
- This simplifies to:
\[ = \hat{i} \left(-\frac{1}{4} + \frac{1}{4}\right) - \hat{j} \left(-\frac{1}{4} - \frac{1}{4}\right) + \hat{k} \left(-\frac{1}{4} + \frac{1}{4}\right) = 0\hat{i} + \hat{j}(-\frac{1}{2}) + 0\hat{k} \]

4. **Final Cross Product:**
- Now calculate \( (\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t}) \):
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & -\frac{1}{2} & 0 \end{vmatrix} \]
- Expanding this determinant gives:
\[ = \hat{i} \left(\frac{1}{2} \cdot 0 - 0 \cdot -\frac{1}{2}\right) - \hat{j} \left(\frac{1}{2} \cdot 0 - 0 \cdot \frac{1}{2}\right) + \hat{k} \left(\frac{1}{2} \cdot -\frac{1}{2} - \frac{1}{2} \cdot 0\right) \]
- This simplifies to:
\[ = 0\hat{i} + 0\hat{j} - \frac{1}{4}\hat{k} = -\frac{1}{4}\hat{k} \]

5. **Magnitude:**
- Finally, calculate the magnitude:
\[ |(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t})| = \left| -\frac{1}{4}\hat{k} \right| = \frac{1}{4} \]

### Final Answer:
The value of \( |(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t})| \) is \( \frac{1}{4} \).