⇒Line and Plane

Correct answer option is (D)

The given lines are $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$

Here,

$\begin{array}{rl}(x_1,y_1,z_1)& \equiv (1,-1,1)\\ (x_2,y_2,z_2)& \equiv (-2,1,-1)\\ (a_1,b_1,c_1)& \equiv (3,2,5)\\ (a_2,b_2,c_2)& \equiv (4,3,2)\end{array}$

$\begin{array}{rl}& \text{ Consider }\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|\\ & =\left|\begin{array}{ccc}-3& 2& -2\\ 3& 2& 5\\ 4& 3& 2\end{array}\right|\\ & =-3(-11)-2(-14)-2(1)\\ & =33+28-2\\ & =59\ne 0\end{array}$

$\therefore$ The lines are not intersecting.

Correct answer option is (A)

Let $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}=\lambda$

$\Rightarrow x=3+\lambda ,y=3\lambda -1,\mathrm{z}=-\lambda +6$

Let $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}=\mu$

$\Rightarrow x=7\mu -5,y=-6\mu +2,z=4\mu +3$

Both the given lines intersect each other.

So, $\lambda +3=7\mu -5$

$\Rightarrow 7\mu -\lambda =8$ .... (i)

Also, $3\lambda -1=-6\mu +2$

$\Rightarrow 6\mu +3\lambda =3$ .... (ii)

From (i) and (ii), we get

$\begin{array}{rl}& \mu =1,\lambda =-1\\ & \text{ i.e., }x=2,y=-4,z=7\end{array}$

$\therefore$ Co-ordinates of the intersection of the given lines are $R(2,-4,7)$

Hence, reflection of $\mathrm{R}$ in the $xy$-plane is $(2,-4,-7)$.

Correct answer option is (D)

Let $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$

Since given lines intersect, we write

$2\lambda +1=\mu +3$ ..... (1)

$3\lambda -1=2\mu +\mathrm{k}$ .... (2)

$4\lambda +1=\mu$ ..... (3)

Substituting value of $\mu$ from (3) in (1), we get

$\begin{array}{rl}& 2\lambda +1=(4\lambda +1)+3\Rightarrow 2\lambda =-3\Rightarrow \lambda =\frac{-3}{2}\\ & \therefore \mu =4\left(\frac{-3}{2}\right)+1=-6+1=-5\end{array}$

Substituting values of $\lambda$ and $\mu$ in (2), we get

$3\left(\frac{-3}{2}\right)-1=2(-5)+k\Rightarrow k=\frac{9}{2}$