Question

Consider matrices $A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 1& 2\\ 1& -1& 1\end{array}\right];B=\left[\begin{array}{ccc}2& 1& 3\\ 4& 1& -1\\ 2& 2& 3\end{array}\right]$; $C=\left[\begin{array}{c}14\\ 12\\ 2\end{array}\right];D=\left[\begin{array}{c}13\\ 11\\ 14\end{array}\right];X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ such that solutions of equation $AX=C$ and $BX=D$ represents two points $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ respectively in three dimensional space. If $P^{\prime }{Q}^{\prime }$ is the reflection of the line $PQ$ in the plane $x+y+z=9$, then the point which does not lie on $P^{\prime }{Q}^{\prime }$ is:

• A. (3,4,2)
• B. (5,3,4)
• C. (7,2,3)
• D. (1,5,6)

Answer 1

Answer: (A)

On solving, $AX=C$ and $BX=D$
$x=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$
$x=\left[\begin{array}{c}3\\ 1\\ 2\end{array}\right]$
$P=(1,2,3),Q=(3,1,2)$
$P{P}^{\prime }:\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}=\lambda$

$(\lambda +1,\lambda +2,\lambda +3)$ lies on plane
$3\lambda +6=9\Rightarrow \lambda =1$
$\therefore P^{\prime }=(3,4,5)$
Similarly $Q^{\prime }=(5,3,4)$
Now check the options.