Given that, equation of planes are x+y+z=1...(i) and 2x+3y−z+4=0...(ii) ∴ The equation of the plane through the intersection of the planes (i) and (ii) is (x+y+z−1)+k(2x+3y−z+4)=0 ...(iii) ⇒x(1+2k)+y(1+3k)+z(1−k)−1(1−4k)=0 ⇒x(1+2k)=0[∵ plane is parallel to X -axis] ⇒1+2k=0⇒k=−
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Put k=−1∕2 in equation (iii), we get the required equation of plane. 2(x+y+z−1)−2x−3y+z−4=0 ⇒2x+2y+2z−2−2x−3y+z−4=0 ⇒−y+3z−6=0 ⇒y−3z+6=0