KEAM 2015 Math Paper

since, plane passes through mid-point of (3,2,6) and (5,4,8)
Hence, (‌

3+5

2

,‌

2+4

2

,‌

6+8

2

) will lie on the plane. Also, plane is perpendicular to the line segment joining (3,2,6) and (5,4,8). Thus, DR's of the normal will be 5−3,4−2,8−6 i.e. 2:2:2 or 1:1:1. Hence, required equation of the plane will be
1(x−4)+1(y−3)+1(z−7)=0
⇒‌‌x+y+z=14