"Let the line $L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$ intersect the plane $2 x+y+3 z=16$ at the point $P$. Let the point $Q$ be the foot of perpendicular from the point $R(1,-1,-3)$ on the line $L$. If $\alpha$ is the area of triangle $P Q R$, then $\alpha^{2}$ is equal to __________."

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"$\quad L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}=r$ (say)

Let $P \equiv\left(2 r_{1}+1,-r_{1}, r_{1}+3\right)$

$P$ lies on $2 x+y+3 z=16$

$\therefore 2\left(2 r_{1}+1\right)+\left(-r_{1}-1\right)+3\left(r_{1}+3\right)=16$

$r_{1}=1$

$P \equiv(3,-2,4)$

$R \equiv(1,-1,-3)$

Let $Q \equiv\left(2 r_{2}+1,-r_{2}-1, r_{2}+3\right)$

$D R$ s of $Q R \equiv\left(2 r_{2}-r_{2} r_{2}+6\right)$

DRs of $L \equiv(2,-1,1)$

$Q R \perp L \Rightarrow 4 r_{2}+r_{2}+r_{2}+6=0$

$r_{2}=-1$

$Q \equiv(-1,0,2)$

$\overrightarrow{Q P} \times \overrightarrow{R P}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & -2 & 2 \\ 2 & -1 & 7\end{array}\right|=-12 \hat{i}-24 \hat{j}+0 \hat{k}$

$\alpha=[P Q R]=\frac{1}{2}|\overrightarrow{Q P} \times \overrightarrow{R P}|=\frac{1}{2} \times 12 \sqrt{5}$

$=6 \sqrt{5}$

$\alpha^{2}=180$"

Let the line $L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$ intersect the plane $2 x+y+3 z=16$ at the point $P$. Let the point $Q$ be the foot of perpendicular from the point $R(1,-1,-3)$ on the line $L$. If $\alpha$ is the area of triangle $P Q R$, then $\alpha^{2}$ is equal to __________.

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Let the line $L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$ intersect the plane $2 x+y+3 z=16$ at the point $P$. Let the point $Q$ be the foot of perpendicular from the point $R(1,-1,-3)$ on the line $L$. If $\alpha$ is the area of triangle $P Q R$, then $\alpha^{2}$ is equal to __________.

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