If \( \vec{F}=2 i+3 j \) acts at point \( C \) and \( \overline{A B}=4 i+6 j \) and \( \overline{M_{A}}=\left(m^{2}+4\right) k, \overline{M_{B}}= \) \( (4 m) k \), then \( m=\cdots \cdots \) \( \begin{array}{llll}\text { (a) } 3 & \text { (b) } 2 & \text { (c) } 4 & \text { (d) } 1\end{array} \)

To determine the value of \( m \), let's analyze the given information step by step. 1. **Given Vectors:** - Force vector: \( \vec{F} = 2\mathbf{i} + 3\mathbf{j} \) acting at point \( C \). - Vector \( \overline{AB} = 4\mathbf{i} + 6\mathbf{j} \). 2. **Moments:** - Moment about point \( A \): \( M_A = (m^2 + 4)\mathbf{k} \). - Moment about point \( B \): \( M_B = 4m\mathbf{k} \). 3. **Equilibrium Condition:** For equilibrium, the total moment about any point should be zero. This implies: \[ M_A = M_B \] Substituting the given moments: \[ m^2 + 4 = 4m \] Rearranging the equation: \[ m^2 - 4m + 4 = 0 \] 4. **Solving the Quadratic Equation:** \[ m = \frac{4 \pm \sqrt{(4)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 16}}{2} = \frac{4 \pm 0}{2} = 2 \] So, \( m = 2 \). **Answer:** Option (b) \( 2 \).