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Let the equation of the plane passing through the line x-2 y-z-5=0=x+y+3 z-5 and parallel to the line x+y+2 z-7=0=2 x+3 y+z-2 be a x+b y+c z=65. Then the distance of the point (a, b, c)  from the plane 2 x+2 y-z+16=0 is 

Option: 1

9


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Equation of plane is

(x-2 y-z-5)+b(x+y+3 z-5)=0
\left|\begin{array}{ccc}1+\mathrm{b} & -2+\mathrm{b} & -1+3 \mathrm{~b} \\ 1 & 1 & 2 \\ 2 & 3 & 1\end{array}\right|=0

\Rightarrow \mathrm{b}=12

Plane is \quad 13 x+10 y+35 z=65
Distance From given point is = 9

Posted by

Anam Khan

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