A force F = 2 i + x j + y k (in N) with y > 0 acts on a particle,...

The work done by the force is zero, so the dot product of force and displacement is zero:
$W = \vec{F} \cdot \vec{s} = 0$
$(2)(1) + (x)(-2) + (y)(2) = 0$
$2 - 2x + 2y = 0 \Rightarrow x - y = 1 \Rightarrow x = y + 1$

The magnitude of the force is $15$ N:
$|\vec{F}| = \sqrt{2^2 + x^2 + y^2} = 15$
$4 + x^2 + y^2 = 225$
$x^2 + y^2 = 221$

Substitute $x = y + 1$ into the magnitude equation:
$(y + 1)^2 + y^2 = 221$
$y^2 + 2y + 1 + y^2 = 221$
$2y^2 + 2y - 220 = 0$
$y^2 + y - 110 = 0$
$(y + 11)(y - 10) = 0$

Since $y > 0$, we have $y = 10$.
Then, $x = 10 + 1 = 11$.
The force vector is $\vec{F} = 2\hat{i} + 11\hat{j} + 10\hat{k}$.

The torque about the origin is given by $\vec{\tau} = \vec{r} \times \vec{F}$:
$\vec{\tau} = (\hat{i} + \hat{j}) \times (2\hat{i} + 11\hat{j} + 10\hat{k})$
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 2 & 11 & 10 \end{vmatrix}$
$\vec{\tau} = \hat{i}(10 - 0) - \hat{j}(10 - 0) + \hat{k}(11 - 2)$
$\vec{\tau} = 10\hat{i} - 10\hat{j} + 9\hat{k}$

The square of the magnitude of the torque is:
$|\vec{\tau}|^2 = 10^2 + (-10)^2 + 9^2 = 100 + 100 + 81 = 281$
Since the magnitude is $\sqrt{N}$, we have $N = 281$.

Answer: $281$