Three Dimensional Geometry Question 9
Question 9 - 25 January - Shift 1
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
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Answer: 9
Solution:
Formula: Equation Of A Plane, A Plane and A Point
Equation of plane is
$ \begin{aligned} & (x-2 y-z-5)+b(x+y+3 z-5)=0 \\ & \begin{vmatrix} 1+b & -2+b & -1+3 b \\ 1 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix} =0 \\ & \Rightarrow b=12 \end{aligned} $
$\therefore$ Plane is $13 x+10 y+35 z=65$
$\therefore$ The distance is 9.
