Question

If the equation of the plane passing through the point $(1,1,2)$ and perpendicular to the line $x-3 y+$ $2 z-1=0=4 x-y+z$ is $A t+ B y+ C z=1$, then $140( C - B + A )$ is equal to_______

Answer 1

The correct answer is 15.
x−3y+2z−1=0
4x−y+z=0
n1​×n2​=∣∣​i^14​j^​−3−1​k^21​∣∣​
=−i^+7j^​+11k^
Dr of normal to the plane is −1,7,11
Equation of plane :
−1(x−1)+7(y−1)+11(z−2)=0
−x+7y+11z=28
28−1​x+287y​+2811z​=1
Ax+By+Cz=1
140(C−B+A)=140(2811​−287​−281​)
=140×283​=15