Find the distance of the point `(-1,-5,-10)` from the point of the intersection of the line `vecr = 2hati-2hatk+lambda(3hati+4hatj+2hatk)` and the plane `vecr.(hati-hatj+hatk) = 5`.

Find the distance of the point `(-1,-5,-10)` from the point of the int

1.	Find the distance of the point `(-1,-5,-10)` from the point of the intersection of the line `vecr = 2hati-2hatk+lambda(3hati+4hatj+2hatk)` and the plane `vecr.(hati-hatj+hatk) = 5`.
Answer» Correct Answer - 13 units. The cartesian equation of the planes is `(x-2)/3=(y+1)/4=(z-2)/2=lambda` (say). A general point on line (i) is `N(3lambda+2,4lambda-1,2lambda+2)`. The Cartesian equation of the given plan is `x-y+z=5`. If this point lies on the given plane, we have `(3lambda+2)-(4lambda-1)+(2lambda+2)=5 rArr lambda=0`. `therefore` point P is `P(2,-1,2)` and the other point is `Q(-1,-5,-10)`. `therefore` PQ =13 units.

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Find the equationof the plane passing through each group of points. i) A(2,2,-1), B(3,4,2) and C(7,0,6) ii) A(0,-1,-1), B(4,5,1) and C(3,9,4) iii) A(-2,6,-6), B(-3,10,-9) and C(-5,0,-6).
Find the coordinates of the foot of the perpendicular and the perpendicular distance from the point P(3,2,1) to the plane `2x-y+z+1=0`.
Find the distance of the point `(-1,-5,-10)` from the point of the intersection of the line `vecr = 2hati-2hatk+lambda(3hati+4hatj+2hatk)` and the plane `vecr.(hati-hatj+hatk) = 5`.
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