Step 1: Write the plane in intercept form.
With intercepts $a=\dfrac{5}{2}$, $b$, $c$ on the axes, the plane is $\dfrac{x}{5/2}+\dfrac{y}{b}+\dfrac{z}{c}=1$, that is $\dfrac{2x}{5}+\dfrac{y}{b}+\dfrac{z}{c}=1$.
Step 2: Use that it passes through $(1,1,1)$.
Substituting gives $\dfrac{2}{5}+\dfrac{1}{b}+\dfrac{1}{c}=1$, so with $u=\dfrac1b$, $v=\dfrac1c$ we have $u+v=\dfrac{3}{5}$.
Step 3: Use the distance from the origin.
The distance from the origin is $\dfrac{1}{\sqrt{(2/5)^2+u^2+v^2}}=\dfrac{5}{7}$. Squaring, $\dfrac{4}{25}+u^2+v^2=\dfrac{49}{25}$, so $u^2+v^2=\dfrac{45}{25}=\dfrac{9}{5}$.
Step 4: Find the product $uv$.
From $(u+v)^2=u^2+v^2+2uv$, we get $\dfrac{9}{25}=\dfrac{9}{5}+2uv=\dfrac{45}{25}+2uv$, so $2uv=-\dfrac{36}{25}$ and $uv=-\dfrac{18}{25}$.
Step 5: Solve the quadratic for $u$ and $v$.
$u,v$ are roots of $t^2-\dfrac{3}{5}t-\dfrac{18}{25}=0$, that is $25t^2-15t-18=0$. Factoring, $(5t-6)(5t+3)=0$, so $t=\dfrac{6}{5}$ or $t=-\dfrac{3}{5}$.
Step 6: Pick the sign that makes $y$ intercept negative.
The $y$ intercept is $b=\dfrac1u$. A negative $y$ intercept needs $u<0$, so $u=-\dfrac{3}{5}$ and $b=-\dfrac{5}{3}$ (then $v=\dfrac{6}{5}$ gives the positive $z$ intercept $c=\dfrac56$, as required).
\[ \boxed{-\dfrac{5}{3}} \]