The image of the line frac{x-1}{3}=frac{y-3}{ | vedclass

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(D) Let the given line be $L: \frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5} = k$. Any point on $L$ is $(3k+1, k+3, -5k+4)$.For the intersection point of

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$\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$

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(D) Let the given line be $L: \frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5} = k$. Any point on $L$ is $(3k+1, k+3, -5k+4)$.For the intersection point of $L$ and the plane $2x-y+z+3=0$,we have $2(3k+1)-(k+3)+(-5k+4)+3=0$.$6k+2-k-3-5k+4+3=0 \implies 6=0$,which is impossible. Thus,the line is parallel to the plane.The image of point $P(1, 3, 4)$ on the line in the plane $2x-y+z+3=0$ is $P'(x', y', z')$.Using the formula $\frac{x'-1}{2} = \frac{y'-3}{-1} = \frac{z'-4}{1} = -2 \frac{2(1)-3+4+3}{2^2+(-1)^2+1^2} = -2 \frac{6}{6} = -2$.$x'-1 = -4 \implies x' = -3$.$y'-3 = 2 \implies y' = 5$.$z'-4 = -2 \implies z' = 2$.The image line passes through $(-3, 5, 2)$ and has the same direction ratios $(3, 1, -5)$ as the original line.Thus,the equation is $\frac{x+3}{3} = \frac{y-5}{1} = \frac{z-2}{-5}$.

The image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2x-y+z+3=0$ is the line

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