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

$ x-3 y+2 z-1=0 $
$ 4 x-y+z=0$
$\overrightarrow{ n }_1 \times \overrightarrow{ n }_2=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & -3 & 2 \\ 4 & -1 & 1\end{vmatrix}$
$=-\hat{ i }+7 \hat{ j }+11 \hat{ k }$
Dr of normal to the plane is $-1,7,11$
Equation of plane :
$ -1(x-1)+7(y-1)+11(z-2)=0$
$ -x+7 y+11 z=28$
$ \frac{-1}{28} x+\frac{7 y}{28}+\frac{11 z}{28}=1$
$ Ax + By + Cz =1 $
$140( C - B + A )=140\left(\frac{11}{28}-\frac{7}{28}-\frac{1}{28}\right)$
$ =140 \times \frac{3}{28}=15$