1. Tardigrade
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  4. Consider matrices A=[1 2 3 4 1 2 1 -1 1] ; B=[2 1 3 4 1 -1 2 2 3]; C=[ 14 12 2]; D=[13 11 14] ; X=[x y z] such that solutions of equation A X=C and B X=D represents two points P(x1, y1, z1) and Q(x2, y2, z2) respectively in three dimensional space. If P' Q' is the reflection of the line P Q in the plane x+y+z=9, then the point which does not lie on P' Q' is:

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Q. Consider matrices $A=\begin{bmatrix}1 & 2 & 3 \\ 4 & 1 & 2 \\ 1 & -1 & 1\end{bmatrix} ; B=\begin{bmatrix}2 & 1 & 3 \\ 4 & 1 & -1 \\ 2 & 2 & 3\end{bmatrix}$; $C=\begin{bmatrix} 14 \\ 12 \\ 2\end{bmatrix}; D=\begin{bmatrix}13 \\ 11 \\ 14\end{bmatrix} ; X=\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ such that solutions of equation $A X=C$ and $B X=D$ 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 $P' Q'$ is the reflection of the line $P Q$ in the plane $x+y+z=9$, then the point which does not lie on $P' Q'$ is:

Three Dimensional Geometry

Solution:

On solving, $A X =C$ and $B X =D$
$x=\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}$
$x=\begin{bmatrix} 3 \\ 1 \\ 2\end{bmatrix}$
$P=(1,2,3), Q=(3,1,2)$
$P P': \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}=\lambda$
image
$(\lambda+1, \lambda+2, \lambda+3)$ 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.