is a(x−1)+b(y−2)+c(z−3)=0 ......(i) where a.1+b.2+c.3=0 i.e., a+2b+3c=0 ........(ii) Since the plane (i) parallel to the line
x
1
=
y
1
=
z
4
∴a.1+b.1+c.4=0 i.e., a+b+4c=0 ...... (iii) From (ii) and (iii) ,
a
8−3
=
b
3−4
=
c
1−2
=k (let) ∴a=5k,b=−k,c=−k On putting the value of a,b and c in equation (i), 5(x−1)−(y−2)−(z−3)=0 ⇒5x−y−z=0 ........(iv) when x=1,y=0 and z=5; then L.H.S. of equation (iv) =5x−y−2 =5×1−0−5=0 =R.H.S. of equation Hence coordinates of the point (1, 0, 5) satisfy the equation plane represented by equations (iv), Therefore the plane passes through the point (1,0,5)