Q.
The perpendicular bisector of the line segment with endpoints (2,3,2) and (−4,1,4) passes through the point (−3,6,1) and has equation of the form ax+3=by−6=cz−1 where a,b and c are relatively prime integers with a>0. The value of abc−(a+b+c) is equal to
The midpoint of the segment is (22−4,23+1,22+4)=(−1,2,3) and the vector between this point and (−3,6,1) is ⟨−1−(−3),2−6,3−1⟩ and ⟨2,−4,2⟩. ∴ the line has parametric equations x=−3+2t,y=6−4t and z=1+2t. Solving each of these equations for t and setting the expressions equal makes the equation. ax+3=by−6=cz−1⇒1x+3=−2y−6=1z−6, so 1⋅(−2)⋅1−(1−2+1)=−2