$\begin{array}{cc}(a)\:x+y+z-1=0=2x-y-z+3\: & (b)\:x+y+z-1=0=x-2y-z+3\: \\ (c)\:x+y+z-1=0=x-y-2z+3\: & (d)\:x+y+z-1=0=2x-y-z-3\end {array}$

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- Projection of a line on a plane is the line of intersection of the given plane and the plane $\perp$ to this plane which contains the given line.

Given line is $\large\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{3}............(i)$

Given plane is $x+y+z-1=0.................(ii)$

$d.r.$ of the normal $(\overrightarrow n)$ to the plane containing (i) is $\perp$ to

normal to the plane $ (ii)= (1,1,1)$ and also is $\perp$ to the line $(i)=(2,1,3)$

$\therefore \overrightarrow n=(1,1,1)\times(2,1,3)=(2,-1,-1)$

Vector $\perp$ to $\overrightarrow a\:\:and\:\:\overrightarrow b$ is $\overrightarrow a\times\overrightarrow b$

$\therefore$ The eqn. of the plane containing $(i)$ and $\perp$ to $(ii)$ is given by

$2x-y-z+d=0.......(iii)$

But since every point on the line $(i)$ satisfies the eqn. of the plane containing it, $(1,2,3)$

satisfies the equation.

$\Rightarrow\:2-2-3+d=0$ $\Rightarrow\:d=3$

$\therefore$ the eqn. of the plane (iii) becomes $2x-y-z+3=0$

As per the definition of projection of the line $(i)$ on the plane $(ii)$ is given by $(ii)$ and $ (iii)$

$i.e.,\:\:x+y+z-1=0=2x-y-z+3$

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