The distance of the point `(1,-5,""9)`from the plane `x-y+z=5`measured along the line `x=y=z`is :(1) `3sqrt(10)`(2) `10sqrt(3)`(3) `(10)/(sqrt(3))`(4) `(20)/3`A. `3sqrt(10)`B. `10sqrt(3)`C. `(10)/(sqrt(3))`D. `(20)/(3)`

The distance of the point `(1,-5,""9)`from the plane `x-y+z=5`measured

1.	The distance of the point `(1,-5,""9)`from the plane `x-y+z=5`measured along the line `x=y=z`is :(1) `3sqrt(10)`(2) `10sqrt(3)`(3) `(10)/(sqrt(3))`(4) `(20)/3`A. `3sqrt(10)`B. `10sqrt(3)`C. `(10)/(sqrt(3))`D. `(20)/(3)`
Answer» Correct Answer - B Equation of line passing through the point (1, -5, 9) and parallel to x=y=z is `(x-1)/(1)=(y+5)/(1)=(z-9)/(1)=lambda" "("say")` Thus, any point on this line is of the form `(lambda+1,lambda-5,lambda+9).` Now, if `P(lambda+1,lambda-5,lambda+9)` is the point of intersection of line and plane, then `(lambda+1)-(lambda-5)+lambda+9=5` `implies" "lambda+15=5implieslambda=-10` `:."Coordinates of point P are "(-9, -15, -1).` Hence, required distance `sqrt((1+9)^(2)+(-5+15)^(2)+(9+1)^(2))` `=sqrt(10^(2)+10^(2)+10^(2))` `=10sqrt(3)`

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