The d.r.s. of the normal to the plane passing through the origin and the line of intersection of the planes $x+2 y+3 z=4$ and $4 x+3 y+2 z=1$ are

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Aicharya · Accepted Answer

Equation of the plane passing through the line of intersection of the given planes, is $(x z+2 y+3 z-4)+\lambda(4 x+3 y+2 z-1)=0$ Since the plane (1) passes through origin, We get $-4-\lambda=0 \quad \Rightarrow \lambda=-4$ Substituting value of $\lambda$ in equation (1), we get $\begin{aligned} & -15 x-10 y-5 z=0 \quad \Rightarrow 3 x+2 y+z=0 \ & 	herefore 	ext { d.r.s. are }(3,2,1) \end{aligned}$

The d.r.s. of the normal to the plane passing through the origin and the line of intersection of the planes…

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