If Q(0, -1, -3) is the image of the point P i | vedclass

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(A) Let $P$ be $(x_1, y_1, z_1)$. Since $Q(0, -1, -3)$ is the image of $P$ in the plane $3x - y + 4z - 2 = 0$,the line $PQ$ is perpendicular to the pl

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$\frac{\sqrt{91}}{2}$

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(A) Let $P$ be $(x_1, y_1, z_1)$. Since $Q(0, -1, -3)$ is the image of $P$ in the plane $3x - y + 4z - 2 = 0$,the line $PQ$ is perpendicular to the plane.Direction ratios of the normal to the plane are $(3, -1, 4)$.The line $PQ$ passes through $Q(0, -1, -3)$ and is parallel to the normal,so its equation is $\frac{x-0}{3} = \frac{y+1}{-1} = \frac{z+3}{4} = k$.Any point on this line is $(3k, -k-1, 4k-3)$. The midpoint $M$ of $PQ$ lies on the plane.$M = (\frac{3k}{2}, \frac{-k-2}{2}, \frac{4k-6}{2})$. Substituting into the plane equation: $3(\frac{3k}{2}) - (\frac{-k-2}{2}) + 4(\frac{4k-6}{2}) = 2$.$9k + k + 2 + 16k - 24 = 4 \implies 26k = 26 \implies k = 1$.Thus,$M = (1.5, -1, 1)$. The vector $\vec{QP} = 2\vec{QM} = 2(1.5, 0, 4) = (3, 0, 8)$.$P = Q + (3, 0, 8) = (3, -1, 5)$.Now,$\vec{QR} = (3-0, -1-(-1), -2-(-3)) = (3, 0, 1)$ and $\vec{QP} = (3, 0, 8)$.Area of $\Delta PQR = \frac{1}{2} |\vec{QP} 	imes \vec{QR}|$.$\vec{QP} 	imes \vec{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 0 & 8 \ 3 & 0 & 1 \end{vmatrix} = \hat{i}(0) - \hat{j}(3-24) + \hat{k}(0) = 21\hat{j}$.Area $= \frac{1}{2} |21\hat{j}| = \frac{21}{2} = 10.5$. Wait,re-evaluating the provided solution logic: The distance from $R$ to line $PQ$ is needed. The area is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. Base $PQ = \sqrt{3^2 + 0^2 + 8^2} = \sqrt{73}$. Height is the perpendicular distance from $R(3, -1, -2)$ to line $PQ$.Using the cross product method: Area $= \frac{1}{2} |\vec{QP} 	imes \vec{QR}| = \frac{21}{2} = 10.5$.

If $Q(0, -1, -3)$ is the image of the point $P$ in the plane $3x - y + 4z = 2$ and $R$ is the point $(3, -1, -2)$,then the area (in square units) of $\Delta PQR$ is

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