A force `vec F = prop hat i + 3 hat j + 6 hat k` is acting at a point `vec r = 2 hat i - 6 hat j - 12 hat k`. The value of `prop` for which angular momentum about origin is conserved is.

Options

A. 1
B. `-1`
C. 2
D. zero

Explanation

To solve the problem, we need to find the value of `prop` (denoted as `α` in the transcript) for which the angular momentum about the origin is conserved. This occurs when the net torque acting on the system is zero.

### Step-by-Step Solution:

1. **Identify the Given Vectors:**
- The force vector is given as \( \vec{F} = \alpha \hat{i} + 3 \hat{j} + 6 \hat{k} \).
- The position vector is given as \( \vec{r} = 2 \hat{i} - 6 \hat{j} - 12 \hat{k} \).

2. **Torque Calculation:**
- The torque \( \vec{\tau} \) is given by the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \):
\[ \vec{\tau} = \vec{r} \times \vec{F} \]

3. **Set Up the Cross Product:**
- Using the determinant form for the cross product:
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -6 & -12 \\ \alpha & 3 & 6 \end{vmatrix} \]

4. **Calculate the Determinant:**
- Expanding the determinant:
\[ \vec{\tau} = \hat{i} \begin{vmatrix} -6 & -12 \\ 3 & 6 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -12 \\ \alpha & 6 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -6 \\ \alpha & 3 \end{vmatrix} \]

5. **Calculate Each Component:**
- For the \( \hat{i} \) component:
\[ = \hat{i}((-6)(6) - (-12)(3)) = \hat{i}(-36 + 36) = 0 \]
- For the \( \hat{j} \) component:
\[ = -\hat{j}((2)(6) - (-12)(\alpha)) = -\hat{j}(12 + 12\alpha) = -12 - 12\alpha \]
- For the \( \hat{k} \) component:
\[ = \hat{k}((2)(3) - (-6)(\alpha)) = \hat{k}(6 + 6\alpha) \]

6. **Combine the Components:**
- Thus, the torque vector is:
\[ \vec{\tau} = 0 \hat{i} + (-12 - 12\alpha) \hat{j} + (6 + 6\alpha) \hat{k} \]

7. **Set Torque to Zero:**
- For angular momentum to be conserved, we need \( \vec{\tau} = 0 \):
\[ -12 - 12\alpha = 0 \quad \text{(for } \hat{j} \text{ component)} \]
\[ 6 + 6\alpha = 0 \quad \text{(for } \hat{k} \text{ component)} \]

8. **Solve for \( \alpha \):**
- From the \( \hat{j} \) component:
\[ -12 - 12\alpha = 0 \implies 12\alpha = -12 \implies \alpha = -1 \]
- From the \( \hat{k} \) component:
\[ 6 + 6\alpha = 0 \implies 6\alpha = -6 \implies \alpha = -1 \]

9. **Conclusion:**
- The value of \( \alpha \) for which angular momentum about the origin is conserved is:
\[ \alpha = -1 \]