# Find the distance

The equation of plane through points A(2, 5, –3),B(–2, –3, 5) and C(5, 3, - 3) is =0 =0

(x-2){0+16}-(y-5){0-24}+(z+3){8+24} = 0

16x-32+24y-120+32z+96=0

16x+24y+32z-56=0

2x+3y+4z-7=0

Now, distance of plane from (7,2,4) = = OR

Given line, And plane, Intersection point of line and plane can be obtained by solving both equations.

Now, substituting the value of in equation of plane. (2+1+2) +λ (3-4+2) = 5

5+λ =5

λ =0

Thus, the position vector of intersecting point is and point of intersection is (2,-1,2).

The required distance = = Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses RELATED QUESTIONS :

Find the equationMathematics - Board Papers

Find the equationMathematics - Board Papers

Find the length oMathematics - Board Papers

Find the coordinaMathematics - Board Papers

Find the equationMathematics - Board Papers

Show that the linMathematics - Board Papers