Equation of the plane containing the given line
== is
A(x−1)+B(y−2)+C (z−3)=0 ......(i)
where
A+5B+4C=0 .......(ii)
Since the point (3,2,0) contains in the plane (i), therefore
2A+0.B−3C=0 .......(iii)
From equations (ii) and (iii),
===k( let
) ⇒A=−15k,B=9k and
C=−10k Putting the value of
A,B and
C in equation (i), we get
−15(x−1)+9(y−2)−10 (z−3)=0 .......(iv)
Now the coordinates of the point (0,-3,1)
satisfy the equation of the plane (iv) as
−15(0−1)+9(−3−2)− 10(1−3) =15−45+20=0 Hence the point (0, –3, 1) contains in the plane.