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Question

A point P moves in the plane :2x3y+6z4=0 such that the area of PAB, where A(2,2,1) and B(1,4,1) is 14 sq. units. If the plane perpendicular to the plane containing the locus of P are 6x+ay+bz+d1=0 and 6x+ay+bz+d2=0, d1>d2, then the value of (d1d2+a+b3) is

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Solution


[Note: A and B lies in the plane ]

Area of PAB
14=12(AB)height12×32+62+22×height=14height=4
Locus of point P is the pair of straight line parallel to AB

Now, the normal of the required plane i.e.
6^i+a^j+b^k is perendicular to AB as well as perpendicular to n of the given plane .
So,
AB=3^i6^j2^k, n=2^i3^j+6^k186a2b=09+3a+b=0(1)123a+6b=04a+2b=(2)
Solving equations (1) and (2), we get
a=2, b=3
The equation of the planes become
6x2y3z+d1=06x2y3z+d2=0
Distance between them is 8 units, so
|d1d2|62+22+32=8d1d2=56 (d1>d2)(d1d2+a+b17)=56233=17

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