For three non-zero real numbers a, b, c if $\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{2a-3b+5c}{k}$ then the value of k is

Let the given equal ratios be equal to a constant proportionality parameter $m$:
$$\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=m$$
This allows us to express $a,b,\text{ and }c$ in terms of $m$:$a=2m$$b=3m$$c=4m$ Now, substitute these expressions into the final ratio term from the problem:
$$\frac{2a-3b+5c}{k}=m$$
$$\frac{2(2m)-3(3m)+5(4m)}{k}=m$$
$$\frac{4m-9m+20m}{k}=m$$
$$\frac{15m}{k}=m$$
Since $m$ is a non-zero value, we can divide both sides by $m$:
$$\frac{15}{k}=1⟹k=15$$