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Question

# Find the equation of the plane through the line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x−y+z=0. Also find the distance of the plane obtained above, from the origin.

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Solution

## Equation of given planes areP1⇒x+y+z−1=0P2⇒2x+3y+4z−5=0Equation of plane through the line of intersection of planes P1,P2 isP1+λP2=0(x+y+z−1)+λ(2x+3y+4z−5)=0(1+2λ)x+(1+3λ)y+(1+4λ)z+(−1−5λ)=0--- (1)Given that plane represented by equation (1) is perpendicular to planex−y+z=0So we use formulaa1a2+b1b2+c1c2=0So, (1+2λ).1+(1+3λ).(−1)+(1+4λ).1+(−1−5λ)=01+2λ−1−3λ+1+4λ=03λ+1=0λ=−13Put λ=−13 in equation (1), we get(1−23)x+(1−1)y+(1−43)z+23=0x3−z3+23=0x−z+2=0 General points on the line:x=2+3λ,y=−4+4λ,z=2+2λThe equation of plane:→r⋅(^i−2^j+^k)=0The point of intersection of the line and the plane :Substituting general point of the line in the equation of plane and finding the particular value of λ.[(2+3λ)^i+(−4+4λ)^j+(2+2λ)^k]⋅(^i−2^j+^k)=0⇒(2+3λ)⋅1+(−4+4λ)(−2)+(2+2λ)⋅1=0⇒12−3λ=0 or λ=4The point of intersection is :(2+3(4),−4+4(4),(2+2(4))=(14,12,10)Distance of this point from (2,12,5) is=√(14−2)2+(12−12)2+(10−5)2=√122+52=13 units.

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