Let $ P $ be a general point on the line :

$$ \frac{x+2}{2} = \frac{y+1}{-1} = \frac{z}{3} = \lambda $$

Therefore, the coordinates of $ P $ can be expressed as :

$$ P = (-2 + 2\lambda, -1 - \lambda, 3\lambda) $$

Now, let $(h, k, w)$ represent the foot of the perpendicular from $ P $ to the plane $ x + y + z = 3 $.

We use the formula for the foot of the perpendicular from a point to a plane :

$$ h = \frac{-1 \cdot (-2 + 2\lambda - 1 - \lambda + 3\lambda - 3)}{1^2 + 1^2 + 1^2} + (-2 + 2\lambda), \\ k = \frac{-1 \cdot (-2 + 2\lambda - 1 - \lambda + 3\lambda - 3)}{1^2 + 1^2 + 1^2} + (-1 - \lambda), \\ w = \frac{-1 \cdot (-2 + 2\lambda - 1 - \lambda + 3\lambda - 3)}{1^2 + 1^2 + 1^2} + 3\lambda $$

Simplifying these, we get :

$$ h = \frac{+2\lambda}{3}, \\ k = \frac{-7\lambda}{3} + 1, \\ w = \frac{5\lambda}{3} + 2 $$

So, we can express $ \lambda $ and the coordinates $ (h, k, w) $ as :

$$ \frac{h}{2} = \frac{k - 1}{-7} = \frac{w - 2}{5} = \frac{\lambda}{3} $$

Therefore, the locus of the foot of the perpendiculars is :

$$ \frac{x}{2} = \frac{y - 1}{-7} = \frac{z - 2}{5} = \frac{\lambda}{3} $$

Thus, the line on which the feet of the perpendiculars lie is represented as :

$$ \frac{x}{2} = \frac{y - 1}{-7} = \frac{z - 2}{5} $$