Only Altitudes?

Let x=15, y=21 and z=35 be the altitudes, p reciprocal of x, q of y, and r of z.
Let H = 1/2 * (p + q+ r). It is well known that reciprocal of area= 4 * $\sqrt{H*(H-p)*(H-q)*(H-r)}$ .
Putting the values, we get 245 * $\sqrt3$ .
So M = 245.

The approach suggested by Ujjwal Rane, is far better so I am requesting him to post his solution.

The product of any altitude and the corresponding base is constant because it is twice the area of the triangle. For convenience, this product can be taken as $105 \textit{x}$ as 105 is the LCM of 15, 21 and 35

Yielding an area of $A = \frac{105 \textit{x}}{2}$ . . . (I)

i.e. $15 \times b_1 = 21 \times b_2 = 35 \times b_3 = 105 \textit{x} \Rightarrow (b_1, b_2, b_3) = (7 \textit{x}, 5 \textit{x}, 3 \textit{x})$

Giving semiperimeter $S = (7 \textit{x} + 5 \textit{x} + 3 \textit{x})/2 = 15 \textit{x}/2$

From Heron's formula $A = \sqrt { S \times (S - 7\textit{x}) \times (S - 5\textit{x}) \times (S - 3\textit{x}) } =\frac {15\sqrt{3} \textit{x}^2}{4}$ . . . . .(II)

Equating (I) and (II) $\Rightarrow \textit{x} = \frac {14\sqrt{3}}{3}$ and area $A = 245 \sqrt{3} \Rightarrow \textbf{M = 245}$