Question

Consider a cube whose faces are given by
$x + y + z = 3 \sqrt{3 \: }$
,
$x +y +z = 2 \sqrt{3}$

,
$4x - 5y + z = \sqrt{42}$
,
$4x - 5y + z = 2 \sqrt{42}$
,
$2x +y - 3z = \sqrt{14}$
,
$2x + y - 3z =2 \sqrt{14}$
and a triangle whose vertices are (2,1,3) ,(1,1,1) ,(3,1,0) ,then the number of point of intersection of cube and triangle is


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Answer 1

Answer:

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