The plane passing through the points (1, 2, 1),
(2, 1, 2) and parallel to the line, 2x = 3y, z = 1
also passes through the point :

Equation of plane passing through (2, 1, 2)<br><br>a(x $-$ 2) + b(y $-$ 1) + c(z $-$ 2) = 0 ......(1)<br><br>As point (1, 2, 1) also passes through the plane, so it satisfy the equation, <br><br>a(1 $-$ 2) + b(2 $-$ 1) + c(1 $-$ 2) = 0<br><br>$\Rightarrow$ $-$a + b $-$ c = 0 ....(2)<br><br>Given line 2x = 3y and z = 1,<br><br>So, symmetric form of the line<br><br>${x \over 3} = {y \over 2} = {{z - 1} \over 0}$<br><br>$\therefore$ Direction ratio of this line is (3, 2, 0) and Direction ration of plane = (a, b, c)<br><br>As plane is parallel to the line so the normal of the plane is perpendicular to the line.<br><br>$\therefore$ Dot product of direction ratio = 0<br><br>3a + 2b + 0(c) = 0 .....(3)<br><br>Equation of plane, <br><br>$$\left| {\matrix{ {x - 2} &amp; {y - 1} &amp; {z - 2} \cr { - 1} &amp; 1 &amp; { - 1} \cr 3 &amp; 2 &amp; 0 \cr } } \right| = 0$$<br><br>$\Rightarrow 3(1 - y + 2 - z) - 2( - x + 2 + z - 2) = 0$<br><br>$\Rightarrow 9 - 3y - 3z + 2x - 2z = 0$<br><br>$\Rightarrow 2x - 3y - 5z + 9 = 0$<br><br>By checking all options you can see ($-$2, 0, 1) satisfy the equation.