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,$$17\lambda(x-1)+2\lambda(y-1)+(-7\lambda)(z+1)=0$$$$\lambda[17x+2y-7z-26]=0$$$$17x+2y-7z=26$$Maths

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