, 2022

Line $L_1$ has equation $y=\frac32x+k$, and thus has slope $\frac32$.

Since $L_2$ is perpendicular to $L_1$, then its slope is $-\frac23$.
Since $L_1$ has equation $y=\frac32x+k$, its $y$-intercept is $k$.

Line $L_2$ has the same $y$-intercept as $L_1$ (which is at $P(0,k)$).

Thus, $L_2$ has slope $-\frac23$ and $y$-intercept $k$, and so it has equation $y=-\frac23x+k$.

Since $L_2$ intersects the $x$-axis at $Q$, the $x$-coordinate of point $Q$ is the $x$-intercept of $L_2$.

Setting $y=0$ in the equation for
$L_2$ and solving for $x$, we get $0=-\frac23x+k$ or $\frac23x=k$, and so $x=\frac{3k}{2}$.

Written in terms of $k$, the $x$-coordinate of point $Q$ is $\frac{3k}{2}$.
From part (b), the coordinates of $P$ are $(0,k)$, and the coordinates of $Q$ are $\left(\frac{3k}{2},0\right)$.

To determine an expression for the area of $\triangle PQR$, we first need to determine
the coordinates of point $R$.

$L_3$ is parallel to $L_1$ and thus has slope $\frac32$ and equation $y=\frac32x+b$, for some $y$-intercept $b$.

Line $L_3$ passes through point $Q\left(\frac{3k}{2},0\right)$, and so
$0=\frac32\left(\frac{3k}{2}\right)+b$ or
$b=-\frac{9k}{4}$.

Therefore the $y$-intercept of $L_3$ is $-\frac{9k}{4}$ and so $R$ has coordinates $\left(0,-\frac{9k}{4}\right)$.

If we call the origin $O(0,0)$,
then the area of $\triangle PQR$ is
given by $\frac12\times PR\times OQ$,
since height $OQ$ is perpendicular to
the base $PR$.

Since $PR=k-\left(-\frac{9k}{4}\right)=\frac{13k}{4}$
and $OQ=\frac{3k}{2}$, then the area
of $\triangle PQR$ is $\frac12\times\frac{13k}{4}\times\frac{3k}{2}=\frac{39k^2}{16}$.

The area of $\triangle PQR$ is 351,
and so $\frac{39k^2}{16}=351$ or
$k^2=\frac{351\times16}{39}$ or $k^2=144$, and so $k=12$ (since $k>0$).