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 `a x+b y+c z=d` be the equation of the plane passing through the point match Column I with Column II. Column I, Column II `a=` , p. 13 `b=` , q. -3 `c=` , r. 1 `d=` , s. -2

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`