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Q.
The projection of the line $\frac{x+1}{1}=\frac{y}{2}=\frac{z-1}{3}$ on the plane $x - 2y + z = 6$ is the line of intersection of this plane with the plane
Three Dimensional Geometry
Solution:
Equation of the plane through $(-1,0,1)$ is
$a (x+1) + b (y - 0 ) + c( z - 1 ) = 0\,\,\,...(i)$
which is parallel to the given line and perpendicular to the given plane
$- a + 2b + 3c = 0 \,\,\,...(ii)$
and $a - 2b + c = 0\,\,\,... (iii)$
From Eqs. (ii) and (iii), we get
c$ = 0,a = 2b$
From Eq. $(i), 2b (x +1) + by =0$
$\Rightarrow 2x +y + 2 = 0$