$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ intersects the co-ordinate axes at points A, B and C respectively. If $△$ PQR has mid points A, B and C then

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foot of normal to $△$ ABC from O is circumcentre of $△$ PQR

incentres of $△ A B C and △ P Q R$ coincide

centroids of $△$ ABC and $△$ PQR coincide

$ar (Δ P Q R) = 2 \sqrt{a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}}$

answer is A, B, C.

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Detailed Solution

$A C ∥ P R and 2 A C = P R$
So, ABPC is a parallelogram comparing the
coordinates of mid-point of diagonals, we get
P(–a,b,c) and Q(a,–b,c) and R(a,b,–c)
Also, AP is median of $△$ ABC and $△$ PQR so
centroids are
Coinciding. The perpendicular bisector of PR is also
perpendicular
to AC. Therefore circumcentre of $△$ PQR is
orthocenter of $△$ ABC
$\begin{array}{l} ar △ P Q R = 4 ar △ A B C \\ = 2 \sqrt{(O A B)^{2} + (O B C)^{2} + (O A C)^{2}} \end{array}$
Where OAB is the area of the projection of ABC on
the plane XOZ etc.