On which of the following lines does the poin | vedclass

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(C) Let the general point on the line $\frac{x - 4}{2} = \frac{y - 5}{2} = \frac{z - 3}{1} = \lambda$ be $(2\lambda + 4, 2\lambda + 5, \lambda + 3)$.S

Vedclass · Accepted Answer

$\frac{x - 1}{1} = \frac{y - 3}{2} = \frac{z + 4}{-5}$

Vedclass · Answer

(C) Let the general point on the line $\frac{x - 4}{2} = \frac{y - 5}{2} = \frac{z - 3}{1} = \lambda$ be $(2\lambda + 4, 2\lambda + 5, \lambda + 3)$.Since this point lies on the plane $x + y + z = 2$,we substitute the coordinates into the plane equation:$(2\lambda + 4) + (2\lambda + 5) + (\lambda + 3) = 2$$5\lambda + 12 = 2$$5\lambda = -10 \implies \lambda = -2$.Substituting $\lambda = -2$ back into the coordinates,we get the point of intersection $P = (2(-2) + 4, 2(-2) + 5, -2 + 3) = (0, 1, 1)$.Now,we check which line passes through the point $(0, 1, 1)$ by substituting $x=0, y=1, z=1$ into the given options.For option $C$: $\frac{0 - 1}{1} = -1$,$\frac{1 - 3}{2} = -1$,$\frac{1 + 4}{-5} = -1$.Since all ratios are equal to $-1$,the point $(0, 1, 1)$ lies on the line given in option $C$.

On which of the following lines does the point of intersection of the line $\frac{x - 4}{2} = \frac{y - 5}{2} = \frac{z - 3}{1}$ and the plane $x + y + z = 2$ lie?

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