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Question

Mathematics Question on Three Dimensional Geometry

The distance of the point (7,−3,−4)(7,-3,-4) from the plane passing through the points (2,−3,1),(−1,1,−2)(2,-3,1),(-1,1,-2) and (3,−4,2)(3,-4,2) is :

A

525 \sqrt{2}

B

424 \sqrt{2}

C

4

D

5

Answer

525 \sqrt{2}

Explanation

Solution

== ∣x−2−34 y+34−5 z−1−34∣=0\begin{vmatrix} x - 2 & -3 & 4 \\\ y + 3 & 4 & -5 \\\ z - 1 & -3 & 4 \end{vmatrix} = 0
x−z−1=0x−z−1=0
Distance of P(7,−3,−4)P(7,−3,−4) from Plane is
d=∣7+4−12∣=52d = \left| \frac{7+4-1}{\sqrt{2}} \right| = 5\sqrt{2}