An algebra problem by Rishi Sharma

It is given that:

$\frac {x}{7} = \frac {y}{11} = \frac {z}{13} = \frac {2x-3y+z} {m}$

$\Rightarrow y = \dfrac {11}{7} x \quad \quad z = \dfrac {13}{7} x$

$\Rightarrow \dfrac {2x-3y+z} {m} = \dfrac {2x-3 (\frac {11}{7}) x+\frac {13}{7} x } {m} = \dfrac {(14-33+13)x}{7m} = \dfrac {-6x}{7m}$

Since $\space \dfrac {2x-3y+z} {m} = \dfrac {x}{7} = \dfrac {-6x}{7m} \quad \Rightarrow m = \boxed {-6}$