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Question

The equation of the plane containing the line $$\cfrac { x+1 }{ -3 } =\cfrac { y-3 }{ 2 } =\cfrac { z+2 }{ 1 } $$ and the point $$(0,7,-7)$$ is


A
x+y+z=1
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B
x+y+z=2
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C
x+y+z=0
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D
None of these
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Solution

The correct option is D $$x+y+z=0$$
The equation of plane containing the line

$$\cfrac { x+1 }{ -3 } =\cfrac { y-3 }{ 2 } =\cfrac { x+2 }{ 1 } $$
$$a(x+1)+b(y-3)+c(z+2)=0......(i)$$
where, $$-3a+2b+c=0....(ii)$$

This passes through $$(0,7,-7)$$
$$\therefore a+4b-5c=0....(iii)\quad $$
From Eqs (ii) and (iii) we get

$$\cfrac { a }{ -14 } =\cfrac { b }{ -14 } =\cfrac { c }{ -14 } \quad \Rightarrow \cfrac { a }{ 1 } =\cfrac { b }{ 1 } =\cfrac { c }{ 1 } $$

Thus, the required plane is $$x+y+z=0$$

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