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Q.
The straight line $\frac{x-x_{0}}{\ell}=\frac{y-y_{0}}{m}=\frac{z-z_{0}}{n}$ is parallel to the plane $a x+b y+c z=d$ then:
Three Dimensional Geometry
Solution:
The direction ratios of given line are $\ell, m , n$ and that of the normal of the given plane are $a , b , c$. Since the line and the plane are mutually parallel, therefore the line and normal to the plane must be mutually perpendicular. Hence, according to the condition of perpendicularity, we obtain $a \ell+ bm + cn =0$