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Write the equation of the plane 2x-3y+5z...

Write the equation of the plane `2x-3y+5z+1=0` in normal from and find its distance from the origin. Find also the distance between from the point (3,1,2).

Text Solution

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The correct Answer is:
`-(2)/(sqrt(38))x+(3)/(sqrt(38))y-(5)/(sqrt(38))z=(1)/(sqrt(38)),(1)/(sqrt(38))`
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