A vector given by $\vec{P} = f(t)\hat{i} + g(t)\hat{j} + \hat{k}$ moves in such a way that it is always parallel to the vector $\vec{Q} = -f''(t)\hat{i} + f'(t)\hat{j} + \hat{k}$. The magnitude of $\vec{P}$ is

Since $\vec{P}$ is parallel to $\vec{Q}$, their corresponding components must be proportional.

$\dfrac{f(t)}{-f''(t)} = \dfrac{g(t)}{f'(t)} = \dfrac{1}{1}$

From the above, we get $f(t) = -f''(t)$ and $g(t) = f'(t)$.

The equation $f''(t) + f(t) = 0$ is a linear second-order differential equation with the general solution:

$f(t) = A \sin t + B \cos t$

Using $g(t) = f'(t)$, we get:

$g(t) = A \cos t - B \sin t$

The magnitude of $\vec{P}$ is given by:

$|\vec{P}| = \sqrt{f(t)^2 + g(t)^2 + 1}$

Substituting the expressions for $f(t)$ and $g(t)$:

$f(t)^2 + g(t)^2 = (A \sin t + B \cos t)^2 + (A \cos t - B \sin t)^2$

$f(t)^2 + g(t)^2 = A^2 \sin^2 t + B^2 \cos^2 t + 2AB \sin t \cos t + A^2 \cos^2 t + B^2 \sin^2 t - 2AB \sin t \cos t$

$f(t)^2 + g(t)^2 = A^2(\sin^2 t + \cos^2 t) + B^2(\cos^2 t + \sin^2 t) = A^2 + B^2$

Therefore, $|\vec{P}| = \sqrt{A^2 + B^2 + 1}$, which is independent of time $t$ and hence a constant.

Answer: constant