Given, equation of $3x-y+4z=2$ and point $P\left(2,4,7\right)$

Now using the formula of image

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=-2\times \frac{\left(a{x}_1+b{y}_1+c{z}_1+d\right)}{a^2+b^2+c^2}$

We get,

$\frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=-\frac{2\left(6-4+28-2\right)}{26}$

$\Rightarrow \frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=-\frac{28}{13}$

$\Rightarrow x=-\frac{58}{13},y=\frac{80}{13},z=-\frac{21}{13}$

Hence, $2a+b+2c=-\frac{116}{13}+\frac{80}{13}-\frac{42}{13}=-6$