If the equation of the plane passing through the point (1,1,2) and perpendicular to the li…

Question

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

Answer: 15

Accepted Answer

Direction of the line of intersection $=(1,-3,2)\times(4,-1,1)=(-1,7,11)$.
$$-1(x-1)+7(y-1)+11(z-2)=0 \implies -x+7y+11z=28.$$
Dividing by $28$: $A=-\dfrac{1}{28},\ B=\dfrac{7}{28},\ C=\dfrac{11}{28}$.
$$140(C-B+A)=140\left(\frac{11-7-1}{28}\right)=140\times\frac{3}{28}=15.$$

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