The point P is the intersection of the straig | vedclass

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(A) The equation of the line $QR$ passing through $Q(2, 3, 5)$ and $R(1, -1, 4)$ is given by $\frac{x-2}{1-2} = \frac{y-3}{-1-3} = \frac{z-5}{4-5}$,wh

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$\frac{1}{\sqrt{2}}$

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(A) The equation of the line $QR$ passing through $Q(2, 3, 5)$ and $R(1, -1, 4)$ is given by $\frac{x-2}{1-2} = \frac{y-3}{-1-3} = \frac{z-5}{4-5}$,which simplifies to $\frac{x-2}{-1} = \frac{y-3}{-4} = \frac{z-5}{-1} = \lambda$. Thus,any point on the line is $P(2-\lambda, 3-4\lambda, 5-\lambda)$. Since $P$ lies on the plane $5x - 4y - z = 1$,we have $5(2-\lambda) - 4(3-4\lambda) - (5-\lambda) = 1$. $10 - 5\lambda - 12 + 16\lambda - 5 + \lambda = 1 \Rightarrow 12\lambda - 7 = 1 \Rightarrow 12\lambda = 8 \Rightarrow \lambda = \frac{2}{3}$. Substituting $\lambda = \frac{2}{3}$,we get $P = (2-\frac{2}{3}, 3-\frac{8}{3}, 5-\frac{2}{3}) = (\frac{4}{3}, \frac{1}{3}, \frac{13}{3})$. Now,for the foot of the perpendicular $S$ from $T(2, 1, 4)$ to $QR$,let $S = (2-\mu, 3-4\mu, 5-\mu)$. The vector $\vec{TS} = (2-\mu-2, 3-4\mu-1, 5-\mu-4) = (-\mu, 2-4\mu, 1-\mu)$. Since $\vec{TS}$ is perpendicular to the line direction vector $\vec{v} = (-1, -4, -1)$,their dot product is zero: $(-\mu)(-1) + (2-4\mu)(-4) + (1-\mu)(-1) = 0$. $\mu - 8 + 16\mu - 1 + \mu = 0 \Rightarrow 18\mu = 9 \Rightarrow \mu = \frac{1}{2}$. Thus,$S = (2-\frac{1}{2}, 3-2, 5-\frac{1}{2}) = (\frac{3}{2}, 1, \frac{9}{2})$. The distance $PS = \sqrt{(\frac{4}{3}-\frac{3}{2})^2 + (\frac{1}{3}-1)^2 + (\frac{13}{3}-\frac{9}{2})^2} = \sqrt{(-\frac{1}{6})^2 + (-\frac{2}{3})^2 + (-\frac{1}{6})^2} = \sqrt{\frac{1}{36} + \frac{4}{9} + \frac{1}{36}} = \sqrt{\frac{2}{36} + \frac{16}{36}} = \sqrt{\frac{18}{36}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

The point $P$ is the intersection of the straight line joining the points $Q(2, 3, 5)$ and $R(1, -1, 4)$ with the plane $5x - 4y - z = 1$. If $S$ is the foot of the perpendicular drawn from the point $T(2, 1, 4)$ to $QR$,then the length of the line segment $PS$ is

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