(15) A force of magnitude \( 2 \sqrt{17} \) force unit acts in the coordinate plane and the equation of its line of action is \( y-4 x=12 \), then the norm of the moment force about the origin point is \( \ldots . \). moment unit. \( \begin{array}{llll}\text { (a) } 24 & \text { (b) } 26 & \text { (c) } 24 \sqrt{17} & \text { (d) } \sqrt{17}+3\end{array} \)

To find the norm of the moment force about the origin, we start by determining the direction of the force. The line of action of the force is given by the equation \( y - 4x = 12 \), which can be rearranged to \( y = 4x + 12 \). Next, we need to find a point on this line and the unit vector of the force. Let's choose \( x = 0 \) to find the point at \( (0, 12) \). The direction of the force vector can be represented as \( \vec{F} = (a, b) \), normalized over the vector direction defined by the line equation, which can be taken as \( (1, 4) \) since the slope is \( 4 \). Calculating the moment \( M \) about the origin, we have: \[ M = r \times F = (x \cdot F_y - y \cdot F_x) \] For our chosen point \( (0, 12) \), \[ M = (0)(b) - (12)(a) = -12a \] Here, \( a = \sqrt{17} \), so the magnitude of the moment is: \[ |M| = 12 \cdot 2\sqrt{17} = 24\sqrt{17} \] Thus, the answer is \( \text{(c) } 24 \sqrt{17} \).