Consider the lines `L_(1): (x-1)/(2)=(y)/(-1)= (z+3)/(1) , L_(2): (x-4)/(1)= (y+3)/(1)= (z+3)/(2) ` and the planes `P_(1)= 7x+y+2z=3, P_(2): 3x+5y-6z=4`. Let `ax+by+cz=d` be the equation of the plane passing through the point of intersection of lines `L_(1) and L_(2)`, and perpendicular to planes `P_(1) and P_(2)`.
Match Column I with Column II.

Answer

` a to r; b to q; c to s; d to p`

Explanation

Plane perependicular to `P_(1) and P_(2)` has direction ratios of normal
`" "|{:(hati,,hatj,,hatk),(7,,1,,2),(3,,5,,-6):}|= -16hati+48hatj+32hatk" "` (i)
For point of intersection of lines
`" "(2lamda_(1)+1, -lamda_(1), lamda_(1)-3)-=(lamda_(2)+4, lamda_(2)-3, 2lamda_(2)-3)`
`rArr" "2lamda_(1)+1= lamda_(2)+ 4 or 2 lamda_(1)-lamda_(2)=3`
and `" "-lamda_(1)= lamda_(2)-3 or lamda_(1)+lamda_(2)=3`
`rArr" "lamda_(1)=2, lamda_(2)=1`
`therefore" "` Point is `(5, -2, -1)" "`(ii)
From (i) and (ii), required planes is
`" "-1(x-5)+ 3(y+2)+ 2(zk+1)=0`
or `" "x -3y-2z=13`
`rArr " "a=1, b=-3, c=-2, d=13`