Find the equation of the plane passing through the point (1, 3, 5) and – InterviewSolution

1.	Find the equation of the plane passing through the point (1, 3, 5) and perpendicular to each of the planes x + 2y + 3z = 7 and 3x + 3y + z = 0, Also find distance of new plane from (2,3,1).
Answer» `p_1=x+2y+3z=7``p_2=3x+3y+z=0``n_1=(1,2,3)``n_2=(3,3,1)`since normal vector is also perpendicular`n2=[[1,j,k],[1,2,3],[3,3,1]] ` on solving, `n_2=-7hat i +8 hat j_3hat` `vec n_3=vec n_1+vec n_2`on solving for `n_3` using (`vec r- vec a)*vec n``-7(x-1)+(y-3)8+(z-5)-3=0` `n3=-7x-8y+3z+2=0`(equation of plane)now the distance;`d=Abs(-5/sqrt(122))`hence solved.

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