Roots and solutions

Yes, since $p$ , $q$ and $r$ are the roots:

$\Rightarrow (x-p)(x-q)(x-r) = x^3 - 7 x^2 + 3x +1$

$x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc = x^3 - 7 x^2 + 3x +1$

$\Rightarrow ab+bc+ca = 3\quad abc = -1$

Since, $\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} = \dfrac {ab+bc+ca} {abc} = \dfrac{3}{-1} = - 3 = - M$

Therefore, $M = \boxed{3}$

1/p+1/q+1/r = pq+qr+pr/pqr = c/-d =-3= -M therefore, M=3

Note that the product of the roots is the NEGATIVE of the constant divided by the first coefficient since the leading degree of power is odd.