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Let the plane $\mathrm{ax} +\mathrm{by}+\mathrm{c} z=\mathrm{d}$ pass through $(2,3,-5)$ and is perpendicular to

the planes $\mathrm{2 x+y-5 z=10}$ and $\mathrm{ 3 x+5 y-7 z=12 \text {. } }$

If $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are integers $\mathrm{d}>0$ and $\operatorname{gcd}(|\mathrm{a}|,|\mathrm{b}|,|\mathrm{c}|, \mathrm{d})=1,$

then the value of $\mathrm{a}+7 \mathrm{~b}+\mathrm{c}+20 \mathrm{~d}$ is equal to :
Option: 1
$18$

Option: 2
$20$

Option: 3
$24$

Option: 4
$22$

Answers (1)

best_answer

DR's of the required plane are given by $\overrightarrow{n_{1}} \times \overrightarrow{n_{2}}$

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

$\mathrm{=(-7+25) i-(-14+15) j+(10-3) k} \\$

$\mathrm{=18 i-j+7 k}$

So, the equation of required plane is

$\mathrm{18 x-y+7 z=2}$

$\mathrm{Using (2,3,-5)}$

$\mathrm{36-3-35=\lambda \Rightarrow \lambda=-2}$

$\mathrm{Plane: 18 x-y+7 z=-2}$

$\mathrm{ \text { But } d>0} \\$

$\mathrm{-18 x+y-7 z=2} \\$

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

$\mathrm{\therefore a+7 b+c+20 d} \\$

$\mathrm{=-18+7-7+40 }\\$

$\mathrm{= 22}$

Hence the correct answer is option 4

Posted by

Shailly goel

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