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JEE MainMathematicsModerate3D

The equation of the line passing through (-4, 3, 1), parallel
to the plane x+2y-z-5=0 and intersecting the
linex+1−3=y−32=z−2−1\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1} is

Show answer and solution
Let the point of intersection with the line be
(−3λ−1,2λ+3,−λ+2)( - 3\lambda - 1,2\lambda + 3, - \lambda + 2)
Hence direction ratio of the line is
(−3λ+3,2λ,−λ+1)( - 3\lambda + 3,2\lambda, - \lambda + 1)
Since this line is parallel to x+2y - z = 5
So v1⋅v2=0    ⇒    −3λ+3+λ+λ−1=0⇒    2λ+2=0⇒λ=−2\begin{matrix} v_{1} \cdot v_{2} = 0\ \ \ \ \Rightarrow \ \ \ \ - 3\lambda + 3 + \lambda + \lambda - 1 = 0 \Rightarrow \ \ \ \ 2\lambda + 2 = 0 \Rightarrow \lambda = - 2 \end{matrix}
Hence direction ratio 6,-2,2 so
linex+43=y−3−1=z−11\frac{x + 4}{3} = \frac{y - 3}{- 1} = \frac{z - 1}{1}

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

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