The cartesian equation of a line passing through $(1,2,3)$ and parallel to planes $x-y+2 z=5$ and $3…

The cartesian equation of a line passing through $(1,2,3)$ and parallel to planes $x-y+2 z=5$ and $3 x+y+z=6$ is
  1. $\frac{x-1}{-3}=\frac{y-2}{-5}=\frac{z-3}{4}$
  2. $\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$
  3. $\frac{x-1}{13}=\frac{y-2}{-1}=\frac{z-3}{1}$
  4. $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{1}$

Solution

Required equation $\begin{aligned} & \frac{x-1}{(-1) \times 1-1 \times 2}=\frac{y-2}{3 \times 2-1 \times 1}=\frac{z-3}{1 \times 1-3 \times(-1)} \\ & \Rightarrow \frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}\end{aligned}$

Asked in: MHT CET 2022 (06 Aug Shift 2)

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