Match the following lists:

List I	List II
a.  Image of the point (3, 5, 7) in the plane 2x+ y + z = - 18 is	p.(- 1, - 1, -1)
b.  The point of intersection of the line $\frac{x-2}{-3}=\frac{y-1}{-2}=\frac{z-3}{2}$ and the plane       2x+y-z=3 is	q. (-21, -7, -5)
c.  The foot of the perpendicular from the point (1, 1,2) to the plane       2x - 2y+42+5=0 is	r. $\left(\frac{5}{2},\frac{2}{3},\frac{8}{3}\right)$
d. The intersection point of the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and       $\frac{x-4}{5}=\frac{y-1}{2}=z$  is	s. $\left(-\frac{1}{12},\frac{25}{12},\frac{-2}{12}\right)$

a.   If the required image is (x, y, z), then $\frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-5}{1}=\frac{\mathrm{z}-7}{1}=-\frac{2(6+5+7+18)}{2^2+1^2+1^2}=-12$ $\text{ or }(x,y,z)\equiv (-21,-7,-5)$

b.  Any point on the line $\frac{x-2}{-3}=\frac{y-1}{2}=\frac{z-3}{2}=\lambda$ is $(-3\lambda +2,2\lambda +1,2\lambda +3)$  which lies on plane 2x + y - z = 3. 

     Therefore 

     $\begin{array}{l}-6\mathrm{\lambda }+4+2\mathrm{\lambda }+1-2\mathrm{\lambda }-3=3\\ -6\mathrm{\lambda }=1\\ \mathrm{\lambda }=-1/6\end{array}$

      $\text{ Therefore, the point is }\left(\frac{5}{2},\frac{2}{3},\frac{8}{3}\right)\text{. }$

c.   If (x, y, z) is the required foot of the perpendicular, then $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-1}{-2}=\frac{\mathrm{z}-2}{4}=-\frac{(2-2+8+5)}{2^2+(-2{)}^2+4^2}$

      $\text{ or }(x,y,z)\equiv \left(\frac{-1}{12},\frac{25}{12},\frac{-2}{12}\right)$

d.    Any point on the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$ is $P(2\lambda +1,3\lambda +2,4\lambda +3)$ which satisfies the line $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{1}$

        $\begin{array}{l}\mathrm{or} \frac{2\mathrm{\lambda }+1-4}{5}=\frac{3\mathrm{\lambda }+2-1}{2}=\frac{4\mathrm{\lambda }+3}{1}\\ \mathrm{or} \mathrm{\lambda }=-1\end{array}$

         The required point is (- 1, - 1, - 1)