If parabola $y^2=\lambda x$ and $25\left[{\left(x-3\right)}^2+{\left(y+2\right)}^2\right]={\left(3x-4y-2\right)}^2$ are equal, then value of $\lambda$ is

Two parabolas are equal if the length of their latusrectum are equal

Length of the latus rectum of $y^2=\lambda x$is $\lambda$.

The equation of the second parabola is

$25\left\{{\left(x-3\right)}^2+{\left(y+2\right)}^2\right\}={\left(3x-4y-2\right)}^2$

$\Rightarrow \sqrt{{\left(x-3\right)}^2+{\left(y+2\right)}^2}=\frac{\left|3x-4y-2\right|}{\sqrt{3^2+4^2}}$

Here focus is $\left(3,-2\right)$and equation of the directrix is $3x-4y-2=0$

Therefore, length of the latus rectum is $2\times$distance between focus and directrix 

$=2\left|\frac{3\times 3-4\times \left(-2\right)-2}{\sqrt{3^2+{\left(-4\right)}^2}}\right|=6$

Thus, the two parabola are equal if $\lambda =6$