Find the coordinates of the point where the line x-1 2 - y-2 3 - z-3 4 meets the plane x-y-4z-6-

The equation of line is x+y+4z=6⋯ (1) The equation of plane is x+12=y+23=z+34=k x=2k 1 y=3k 2 z=4k 3 2k 1+3k 2+4(4k 3)=6 21k 15=6 ...

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Find the coordinates of the point where the line $\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z + 3}{4}$ meets the plane $x + y + 4 z = 6.$

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The point where the line $\frac{x - 1}{2} = \frac{y - 2}{- 3} = \frac{z + 3}{4}$ meets the plane 2x + 4y - z = 1, is [DSSE 1981]

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

x - y + z - 5 = 0

x = 2, y - z = 0

x + z = 3.

\frac{x}{2} \equiv \frac{y}{3} = \frac{z}{1}

\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{- 1}

\frac{x + 5}{7} = \frac{y - 2}{- 6} = \frac{z - 3}{4}

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Find the coordinates of the point where the line through (3, −4, −5) and (2, − 3, 1) crosses the plane 2x + y + z = 7).

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