If two distinct points Q and R lie on the lin | vedclass

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(B) The line of intersection of the planes $-x + 2y - z = 0$ and $3x - 5y + 2z = 0$ has a direction vector $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j

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$\frac{4}{3} \sqrt{38}$

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(B) The line of intersection of the planes $-x + 2y - z = 0$ and $3x - 5y + 2z = 0$ has a direction vector $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 2 & -1 \ 3 & -5 & 2 \end{vmatrix} = \hat{i}(4-5) - \hat{j}(-2+3) + \hat{k}(5-6) = -\hat{i} - \hat{j} - \hat{k}$.Thus,the direction ratios of the line are $(1, 1, 1)$.Let $T$ be the projection of $P(1, -2, 3)$ onto the line. The coordinates of any point on the line are $(\alpha, \alpha, \alpha)$.The vector $\vec{PT} = (\alpha - 1, \alpha + 2, \alpha - 3)$.Since $\vec{PT}$ is perpendicular to the line $(1, 1, 1)$,we have $1(\alpha - 1) + 1(\alpha + 2) + 1(\alpha - 3) = 0$,which gives $3\alpha - 2 = 0$,so $\alpha = \frac{2}{3}$.The point $T$ is $(\frac{2}{3}, \frac{2}{3}, \frac{2}{3})$.$PT^2 = (\frac{2}{3} - 1)^2 + (\frac{2}{3} + 2)^2 + (\frac{2}{3} - 3)^2 = (-\frac{1}{3})^2 + (\frac{8}{3})^2 + (-\frac{7}{3})^2 = \frac{1 + 64 + 49}{9} = \frac{114}{9} = \frac{38}{3}$.In $	riangle PQT$,$\cos 	heta = \frac{PT}{PQ} = \frac{\sqrt{38/3}}{\sqrt{18}} = \sqrt{\frac{38}{3 	imes 18}} = \sqrt{\frac{19}{27}}$.Then $\sin 	heta = \sqrt{1 - \frac{19}{27}} = \sqrt{\frac{8}{27}} = \frac{2\sqrt{2}}{3\sqrt{3}}$.The area of $	riangle PQR = 2 	imes 	ext{Area}(	riangle PQT) = 2 	imes (\frac{1}{2} 	imes PT 	imes QT) = PT 	imes (PQ \sin 	heta) = \sqrt{\frac{38}{3}} 	imes \sqrt{18} 	imes \frac{2\sqrt{2}}{3\sqrt{3}} = \sqrt{\frac{38}{3}} 	imes 3\sqrt{2} 	imes \frac{2\sqrt{2}}{3\sqrt{3}} = \sqrt{38} 	imes \frac{2 	imes 2}{3} = \frac{4}{3} \sqrt{38}$.

If two distinct points $Q$ and $R$ lie on the line of intersection of the planes $-x + 2y - z = 0$ and $3x - 5y + 2z = 0$,and $PQ = PR = \sqrt{18}$,where the point $P$ is $(1, -2, 3)$,then the area of the triangle $PQR$ is equal to

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