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Question

Find the equation of the plane passing through the point $$(1,1-1)$$ and perpendicular to the plnes $$x+2y+3z-7=0$$ and $$2x-3y+4z=0$$.


Solution

Let the DR's of the normal to the required plane be $$a,b,c$$.
Since the required plane passes through the point $$(1,1,-1)$$.
So, its equation is :
$$a(x-1)+b(y-1)+c(z+1)=0$$       ........(1)
Also, plane 1 is perpendicular to each of the given planes. Therefore,
$$a+2b+3c=0$$            .......(2)
and $$2a-3b+4c=0$$           .........(3)
On solving equation 2 and 3, we get,
$$\dfrac{a}{17}=\dfrac{b}{2}=\dfrac{c}{-7}=\lambda$$

$$a=17\lambda, b=2\lambda,c=-7\lambda$$
On putting the values of a, b, c in equation 1, we get,
$$17\lambda(x-1)+2\lambda(y-1)+(-7\lambda)(z+1)=0$$
$$\lambda[17x+2y-7z-26]=0$$
$$17x+2y-7z=26$$

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