Let A(0, 6, 8) and B(6, 12, 0) be two given points and $P (a, 0, 0)$ be a point on x–axis such that PA+PB is minimum, then which of the following is/are true?

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Perpendicular distance from the point A to the plane containing the points P, Q, B is 8 units

$I f p o i n t Q i s o n t h e y - a x i s s u c h t h a t Q A + Q B i s m i n i m u m t h e n a r e a o f Δ P O Q i s \frac{90}{7} u n i t^{2} w h e r e O i s o r i g i n$

The volume of tetrahedron OPAB (where O is origin) is $\frac{240}{11}$ $u n i t^{3}$

The volume of tetrahedron OPAB (where O is origin) is $\frac{480}{11} {unit}^{3}$

answer is A, C, D.

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Detailed Solution

$α = \frac{30}{11}, Q (0, \frac{66}{7}, 0)$

PA+PB is minimum, then So, $\frac{M P}{P N} = \frac{A M}{B N} = \frac{5}{6}$

So, $P = (\frac{30}{11}, 0, 0)$ Similarly Q is $(0, \frac{66}{7}, 0)$

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