Let the line L: frac{x-1}{2} = frac{y+1}{-1} | vedclass

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(A) Any point on line $L$ is given by $(2\lambda+1, -\lambda-1, \lambda+3)$.Substituting this into the plane equation $2x+y+3z=16$:$2(2\lambda+1) + (-

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$180$

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(A) Any point on line $L$ is given by $(2\lambda+1, -\lambda-1, \lambda+3)$.Substituting this into the plane equation $2x+y+3z=16$:$2(2\lambda+1) + (-\lambda-1) + 3(\lambda+3) = 16$$4\lambda + 2 - \lambda - 1 + 3\lambda + 9 = 16$$6\lambda + 10 = 16 \Rightarrow 6\lambda = 6 \Rightarrow \lambda = 1$.Thus,point $P = (3, -2, 4)$.For the foot of the perpendicular $Q$ from $R(1, -1, -3)$ to line $L$,let $Q = (2\mu+1, -\mu-1, \mu+3)$.The vector $\vec{RQ} = (2\mu, -\mu, \mu+6)$. Since $\vec{RQ}$ is perpendicular to the direction of $L$,$\vec{v} = \langle 2, -1, 1 angle$:$2(2\mu) - 1(-\mu) + 1(\mu+6) = 0$$4\mu + \mu + \mu + 6 = 0 \Rightarrow 6\mu = -6 \Rightarrow \mu = -1$.Thus,$Q = (-1, 0, 2)$.Now,$\vec{QR} = R - Q = (1 - (-1), -1 - 0, -3 - 2) = (2, -1, -5)$.And $\vec{QP} = P - Q = (3 - (-1), -2 - 0, 4 - 2) = (4, -2, 2)$.The cross product $\vec{QR} 	imes \vec{QP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & -5 \ 4 & -2 & 2 \end{vmatrix} = \hat{i}(-2-10) - \hat{j}(4+20) + \hat{k}(-4+4) = -12\hat{i} - 24\hat{j}$.The area $\alpha = \frac{1}{2} |\vec{QR} 	imes \vec{QP}| = \frac{1}{2} \sqrt{(-12)^2 + (-24)^2} = \frac{1}{2} \sqrt{144 + 576} = \frac{1}{2} \sqrt{720}$.Therefore,$\alpha^2 = \frac{1}{4} 	imes 720 = 180$.

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

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