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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 common point of intersection

(B) intersect at a unique point

(C) intersect along a straight line

(D) intersect along a plane
asked in Linear Algebra by Boss (11.3k points) | 76 views

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