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Let $P_1$, $P_2$ and $P_3$ denote, respectively, the planes defined by

$$a_1 x+b_1 y+c_1 z= \alpha_1$$

$$a_2 x+b_2 y+c_2 z= \alpha_2$$

$$a_3 x+b_3 y+c_3 z= \alpha_3$$

It is given that $P_1$, $P_2$ and $P_3$ intersect exactly at one point when $\alpha_1=\alpha_2=\alpha_3=1$. If now $\alpha_1=2$, $\alpha_2=3$ and $\alpha_3=4$ then the planes

 A) do not have any point of intersection. 
 B) intersect at a unique point. 
 C) intersect along a straight line.
 D) intersect along a plane.
erfink
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Gopesh Thakur
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  • Changing the $\alpha_i$ doesn't change the orientation of the planes (the way they're facing), only their specific positioning. What effect would that have on the intersection of the planes? (You might think about the corresponding problem in two dimensions.) – Brian Tung May 10 '17 at 19:26
  • Hint: If there is exactly one solution when $\alpha_1=\alpha_2=\alpha_3$, what does this tell you about the rank of the coefficient matrix? – amd May 10 '17 at 23:22

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