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Find the equation of the plane which ...

Find the equation of the plane which passes through the point `(1,2,3)` and which is at the minimum distance from the point `(-1,0,2)dot`

Text Solution

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The plane passes through the point `A(1, 2, 3)` and is at the maximum distance from the point `B(-1, 0, 2)`, then the plane is perpendicular to line `AB`. Therefore, the direction ratios of the normal to the plane are 2, 2 and 1.
Hence, the equation of the plane is
`" "2(x-1)+2(y-2)+1(z-3)=0`
or `" "2x+2y+z=9`
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