To solve the problem of calculating the work done by a force acting on a particle as it moves from point \( (1, 2, 3) \) to point \( (-1, -2, 3) \), we can follow these steps:
### Step 1: Identify the Force and Displacement
The force acting on the particle is a constant force of \( 10 \, \text{N} \) directed towards the origin. The initial position of the particle is \( \mathbf{A} = (1, 2, 3) \) and the final position is \( \mathbf{B} = (-1, -2, 3) \).
### Step 2: Calculate the Displacement Vector
The displacement vector \( \mathbf{D} \) can be calculated as:
\[
\mathbf{D} = \mathbf{B} - \mathbf{A} = (-1, -2, 3) - (1, 2, 3) = (-1 - 1, -2 - 2, 3 - 3) = (-2, -4, 0)
\]
### Step 3: Calculate the Work Done
The work done \( W \) by a constant force is given by the formula:
\[
W = \mathbf{F} \cdot \mathbf{D}
\]
where \( \mathbf{F} \) is the force vector and \( \mathbf{D} \) is the displacement vector. The force vector \( \mathbf{F} \) can be expressed as:
\[
\mathbf{F} = -10 \hat{r}
\]
where \( \hat{r} \) is the unit vector pointing from the origin to the particle's position.
To find \( \hat{r} \) at the initial position \( (1, 2, 3) \):
\[
\hat{r} = \frac{(1, 2, 3)}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{(1, 2, 3)}{\sqrt{14}}
\]
Thus, the force vector \( \mathbf{F} \) at the initial position is:
\[
\mathbf{F} = -10 \cdot \frac{(1, 2, 3)}{\sqrt{14}} = \left(-\frac{10}{\sqrt{14}}, -\frac{20}{\sqrt{14}}, -\frac{30}{\sqrt{14}}\right)
\]
### Step 4: Calculate the Dot Product
Now we calculate the dot product \( \mathbf{F} \cdot \mathbf{D} \):
\[
W = \left(-\frac{10}{\sqrt{14}}, -\frac{20}{\sqrt{14}}, -\frac{30}{\sqrt{14}}\right) \cdot (-2, -4, 0)
\]
Calculating this gives:
\[
W = \left(-\frac{10}{\sqrt{14}} \cdot -2\right) + \left(-\frac{20}{\sqrt{14}} \cdot -4\right) + \left(-\frac{30}{\sqrt{14}} \cdot 0\right)
\]
\[
W = \frac{20}{\sqrt{14}} + \frac{80}{\sqrt{14}} + 0 = \frac{100}{\sqrt{14}}
\]
### Step 5: Determine the Work Done
Since the force is always directed towards the origin and the displacement is also directed towards the origin, the work done by the force is positive. However, we also need to consider the net work done over the entire path, which can be simplified to zero since the force acts in the opposite direction to the displacement.
Thus, the total work done by the force as the particle moves from \( (1, 2, 3) \) to \( (-1, -2, 3) \) is:
\[
W = 0
\]
### Final Answer
The work done by the force on the particle during its displacement is \( 0 \, \text{J} \).
