29. A line M through A is drawn parallel to BD . Point S moves such that its distances from the line BD and the vertex A are equal. If locus of $S$ cuts $M$ at $T _ { 2 }$ and $T _ { 3 }$ and $A C$ at $T _ { 1 }$, then area of $\Delta T _ { 1 } T _ { 2 } T _ { 3 }$ is\\
(A) $\frac { 1 } { 2 }$ sq. units\\
(B) $\frac { 2 } { 3 }$ sq. units\\
(C) 1 sq. unit\\
(D) 2 sq. units

Sol. (C)\\
$\because \mathrm { AG } = \sqrt { 2 }$\\
$\therefore \mathrm { AT } _ { 1 } = \mathrm { T } _ { 1 } \mathrm { G } = \frac { 1 } { \sqrt { 2 } } \quad \left[ \right.$ as A is the focus, $\mathrm { T } _ { 1 }$ is the vertex and BD is the directrix of parabola].\\
Also $\mathrm { T } _ { 2 } \mathrm {~T} _ { 3 }$ is latus rectum $\therefore \mathrm { T } _ { 2 } \mathrm {~T} _ { 3 } = 4 \times \frac { 1 } { \sqrt { 2 } }$\\
∴ Area of $\Delta \mathrm { T } _ { 1 } \mathrm {~T} _ { 2 } \mathrm {~T} _ { 3 } = \frac { 1 } { 2 } \times \frac { 1 } { \sqrt { 2 } } \times \frac { 4 } { \sqrt { 2 } } = 1$.\\
\includegraphics[max width=\textwidth, alt={}, center]{c8be114f-ece8-451f-b2d5-ce85134c9605-09_343_424_1310_1052}

\section*{Comprehension IV}
$A = \left[ \begin{array} { l l l } 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array} \right]$, if $U _ { 1 } , U _ { 2 }$ and $U _ { 3 }$ are columns matrices satisfying.\\
$\mathrm { AU } _ { 1 } = \left[ \begin{array} { l } 1 \\ 0 \\ 0 \end{array} \right] , \mathrm { AU } _ { 2 } = \left[ \begin{array} { l } 2 \\ 3 \\ 0 \end{array} \right] , \quad \mathrm { AU } _ { 3 } = \left[ \begin{array} { l } 2 \\ 3 \\ 1 \end{array} \right]$ and U is $3 \times 3$ matrix whose columns are $\mathrm { U } _ { 1 } , \mathrm { U } _ { 2 } , \mathrm { U } _ { 3 }$ then answer the following questions\\