Let the plane ax + by + cz = d pass through (2, 3, -5) and is perpendicular to the planes $2\mathrm{x}+\mathrm{y}-5\mathrm{z}=10$ $3\mathrm{x}+5\mathrm{y}-7\mathrm{z}=12$. If a, b, c, d are integers d > 0 and gcd ($\left|\mathrm{a}\right|,\left|\mathrm{b}\right|,\left|\mathrm{c}\right|,\mathrm{d})=1$, then the value of $\mathrm{a}+7\mathrm{b}+\mathrm{c}+20\mathrm{d}$ is equal to :

Given  plane equation is  ax + by + c z = d

It is perpendicular to the planes  $2\mathrm{x}+\mathrm{y}-5\mathrm{z}=10, 3\mathrm{x}+5\mathrm{y}-7\mathrm{z}=12$

DR’S normal of plane

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

$\therefore$ eqⁿ of plane

$18\mathrm{x}-\mathrm{y}+7\mathrm{z}=\mathrm{d}$

It passes through (2, 3, –5)

36 – 3 – 35 = d            $\therefore$d = –2

$\therefore$ Eqⁿ of plane

$\begin{array}{l}18\mathrm{x}-\mathrm{y}+7\mathrm{z}=-2\\ -18\mathrm{x}+\mathrm{y}-7\mathrm{z}=2\\ \therefore \mathrm{a}=-18, \mathrm{b}=1, \mathrm{c}=-7, \mathrm{d}=2\\ \mathrm{a}+7\mathrm{b}+\mathrm{c}+20\mathrm{d}=-18+7-7+40=22\end{array}$