Let the plane $ax+by+cz=d$ pass through $\left(2,3,-5\right)$ and is perpendicular to the planes $2x+y-5z=10$ and $3x+5y-7z=12$

If $a,b,c,d$ are integers $d>0$ and $gcd\left(\left|a\right|,\left|b\right|,\left|c\right|,d\right)=1$ then the value of $a+7b+c+20d$ is equal to

We know that D.R of normal of plane will be given by cross product of given two planes, as it is perpendicular to both given planes, so

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -5\\ 3& 5& -7\end{array}\right|=18\hat{i}-\hat{j}+7\hat{k}$

So, equation of plane is, $18x-y+7z=d$

And it passes through $\left(2,3,-5\right)$

So, $36-3-35=d \therefore d=-2$

So equation of plane is $18x-y+7z=-2$

$\Rightarrow -18x+y-7z=2$

$\therefore a=-18, b=1,c=-7, d=2$

$a+7b+c+20d=-18+7-7+40=22$