Consider the following three planes 

$\begin{array}{l}{p}_1:2x+y+z=1\\ p_2:x-y+z=2\\ p_3:kx-y+3z=5\end{array}$

The three planes interest each other at point R on XOY plane and at the point S on YOZ plane ‘O’ is Origin. If the coordinates of point ‘P’ situated at minimum distance from point ‘O’ on the RS are $(\alpha ,\beta ,\gamma )$. Then the value of $\frac{\frac{\alpha }{2}-180\beta +2\gamma }{100}$ is a three digit number abc and a + b – c is equal to

$\frac{\frac{\alpha }{2}-180\beta +2\gamma }{100}$

$$\begin{gathered} \Rightarrow 2(-3+1)-1(3-k)+1(-1+k)=0 \\ \Rightarrow k=4 \end{gathered}$$

Now put  $z=0$ in  $p_1$ and $p_2$ we get 
 $2x+y=1$ and  $x-y=2$
Solving these we get  point  $R(1,-1,0)$
Again putting x = 0  in p1 and p2  we get
$y+z=1$, and $-y+z=2$

 Solving these ‘2’ equation, we get  $S(0,\frac{1}{2},\frac{3}{2})$
Now Equation of the line $\overleftrightarrow{RS}$ is $\frac{x-1}{1}=\frac{y+}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{z}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\lambda$

$P(-\lambda +1,\frac{\lambda }{2}-1,\frac{3\lambda }{2})$  be any point on  $1=0$
And OR $\perp$ PQ  So, $(-1)(\lambda +1)+\left(\frac{1}{2}\right)\left(\frac{\lambda -2}{2}\right)+\left(\frac{3}{2}\right)\frac{3\lambda }{2}=0$
$\Rightarrow \lambda =\frac{3}{7}\therefore P=(\frac{4}{7},\frac{-11}{14},\frac{9}{14})$

$$\begin{gathered} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\therefore \alpha =\frac{4}{7},\beta =\frac{-11}{14},\gamma =\frac{9}{14}\therefore \frac{\alpha }{2}-180\beta +2k \\ =(\frac{4}{14}+\frac{1980}{14}+\frac{18}{14})=\frac{2002}{14} \\ = 143 \end{gathered}$$

Which is equal to a three digit number abc then a + b – c = 2