Match the Column I with Column II

	Column-I		Column-II
P	The coordinates of a point on the line $x=4y+5,z=3y-6$ at a distance 3 from the point $(5,3,-6)$ is/are	1	$(-1,-2,0)$
Q	The plane containing the lines $\frac{x-2}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and parallel to $\widehat{i}+4\widehat{j}+7\widehat{k}$  has the point	2	$(5,0,-6)$
R	A line passes through two points $A(2,-3,-1)$ and  $B(8,-1,2)$. The coordinates of a point on this line nearer to the origin and at a distance of 14 units from A is/are	3	$(2,5,7)$
S	The coordinates of the foot of the perpendicular from the point  $(3,-1,11)$ on the line $\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ is/are	4	$(-10,-7,-7)$

1)  $a\to q;b\to p;c\to s;d\to r$
a. The given line is x=4y+5,z=3y-6
or  $\frac{x-5}{4}=y,\frac{z+6}{3}=y$
or  $\frac{x-5}{4}=\frac{y}{1}=\frac{z+6}{3}=\lambda$
(say)
Any point on the line is of the form  $(4\lambda +5,\lambda 3\lambda -6)$.
The distance between  $(4\lambda +5,\lambda ,3\lambda -6)$ and $(5,3,-6)$ is 3 units (given). Therefore ${(4\lambda +5-5)}^2+{(\lambda -3)}^2+{(3\lambda -6+6)}^2=9$
or  $16{\lambda }^2+{\lambda }^2+9-6\lambda +9{\lambda }^2=9$

or  $26{\lambda }^2-6\lambda =0$

or $\lambda =0,3/13$

The point is $(5,0,-6)$.
2)  The equation of the plane containing the lines  $\frac{x-2}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and parallel to  $\widehat{i}+4\widehat{j}+7\widehat{k}$ is  $\left|\begin{array}{ccc}x-2& y+3& z+5\\ 1& 4& 7\\ 3& 5& 7\end{array}\right|=0$

or   $x-2y+z-3=0$
Point  $(-1,-2,0)$ lies on this plane.

3) The line passing through points $A(2,-3,-1)$ and $B(8,-1,2)$ is  $\frac{x-2}{8-2}=\frac{y+3}{-1+3}=\frac{z+1}{2+1}$ or  $\frac{x-2}{6}=\frac{y+3}{2}=\frac{z+1}{3}=\lambda$ (say).
Any point on this line is of the form $P(6\lambda +2$,  $2\lambda -3,3\lambda -1)$, whose distance from point $A(2,-3,-1)$ is 14 units. Therefore, $PA=14$
or  $P{A}^2={\left(14\right)}^2$

$\begin{array}{ll}\Rightarrow & {\left(6\lambda \right)}^2+{\left(2\lambda \right)}^2+{\left(3\lambda \right)}^2=196\\ or & 49{\lambda }^2=196\\ or & {\lambda }^2=4\\ or & \lambda =\pm 2\end{array}$

Therefore, the required points are  $(14,1,5)$ and $(-10,-7-7)$. The point nearer to the origin is  $(14,1,5)$.
4)  Any point on line  $AB, \frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$ is  $M(2\lambda ,3\lambda +2,4\lambda +3)$. Therefore, the direction ratios of $PM$  are $2\lambda -3,3\lambda +3$  and  $4\lambda -8$.

But  $PM\perp AB$

$\begin{array}{ll}\therefore & 2(2\lambda -3)+3(3\lambda +3)+4(4\lambda -8)=0\\ & 4\lambda -6+9\lambda +9+16\lambda -32=0\\ & 29\lambda -29=0\end{array}$