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Question

Consider a plane $$x+y-z=1$$ and point $$A(1, 2, -3)$$. A line $$L$$ has the equation $$x=1+3r, y=2-r$$ and $$z=3+4r$$.
The coordinate of a point $$B$$ of line $$L$$ such that $$AB$$ is parallel to the plane is:


A
(10,1,15)
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B
(5,4,5)
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C
(4,1,7)
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D
(8,5,9)
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Solution

The correct option is D $$(-8, 5, -9)$$
Let $$\vec { OB } =\left( 1+3r \right) i+\left( 2-r \right) j+\left( 3+4r \right) k$$
$$\vec { AB } =\vec { OB } -\vec { OA } =\left( 1+3r \right) i+\left( 2-r \right) j+\left( 3+4r \right) k-i-2j+3k=3ri-rj+\left( 6+4r \right) k$$
Since, $$\vec { AB }$$ is parallel to $$x+y-z=1$$
Therefore, $$\vec { AB } .\left( i+j-k \right) =0$$
$$\Rightarrow \left( 3ri-rj+\left( 6+4r \right) k \right) .\left( i+j-k \right) $$
$$\Rightarrow 3r-r-6-4r=0$$
$$\Rightarrow r=-3$$
Therefore, $$\vec { OB } =-8i+5j-9k$$
Ans: D

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