What is the equation of the plane passing through the line 3x−y−5z=2=x−2y+3z and perpendicular to the plane x−y+z=3?
Equation of the plane passing through the line 3x−y−5z=2=x−2y+3z is
3x−y−5z−2+k(x−2y+3z−2)=0
⇒(3+k)x−(1+2k)y+(−5+3k)z=2+2k
Since, it is perpendicular to x−y+z=3
Therefore, ((3+k)i−(1+2k)j+(−5+3k)k).(i−j+k)=0
⇒3+k+1+2k−5+3k=0 ⇒k=16
Therefore, desired plane is ⇒(3+16)x−(1+2(16))y+(−5+3(16))z=2+2(16)
⇒19x−8y−27=14