Question

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

• A. $(0,-6,2)$
• B. $(0,6,-2)$
• C. $(-2,0,1)$
• D. $(2,0,-1)$

Answer 1

The correct option is C

$(-2,0,1)$

Explanation for the correct option:

Find the equation of the plane

Plane passes through $(2,1,2)$ is

$a(x-2)+b(y-1)+(z-2)=0$

It also passes through $(1,2,1)$

$$\begin{gathered} \therefore -a+b-c=o \\ \Rightarrow a-b+c=0.....\left(\mathrm{i}\right) \end{gathered}$$

Given line $2x=3y,z=1$

$\Rightarrow \frac{x}{3}=\frac{y}{2}=\frac{z-1}{0}$ is parallel to $\left(\mathrm{i}\right)$

$$\begin{gathered} \therefore 3a+2b+0c=0 \\ \Rightarrow \frac{a}{0-2}=\frac{b}{3-0}=\frac{c}{2+3} \\ \Rightarrow \frac{a}{-2}=\frac{b}{3}=\frac{c}{5} \\ \Rightarrow \frac{a}{2}=\frac{b}{-3}=\frac{c}{-5} \end{gathered}$$

$\therefore$Plane is $2x-4-3y+3-5z+10=0$

$\Rightarrow 2x-3y-5z=-9.....\left(\mathrm{ii}\right)$

Option (c): By substituting $(x,y,z)=(-2,0,1)$ in the L.H.S. of the equation (ii) we get,

$$\begin{gathered} L.H.S.=2(-2)-3\left(0\right)-5\left(1\right) \\ =-4-0-5 \\ =-9 \\ =R.H.S \end{gathered}$$

Hence, option (C) is correct.

Explanation for the incorrect options:

Option (a): By substituting $(x,y,z)=(0,-6,2)$ in the L.H.S. of the equation (ii) we get,

$$\begin{gathered} L.H.S.=2\left(0\right)-3(-6)-5\left(2\right) \\ =0+18-10 \\ =8 \\ \ne R.H.S \end{gathered}$$

Hence, option (a) is incorrect.

Option (b): By substituting $(x,y,z)=(0,6,-2)$ in the L.H.S. of the equation (ii) we get,

$$\begin{gathered} L.H.S.=2\left(0\right)-3\left(6\right)-5(-2) \\ =0-18+10 \\ =-8 \\ \ne R.H.S \end{gathered}$$

Hence, option (b) is also incorrect.

Option (d): By substituting $(x,y,z)=(2,0,-1)$ in the L.H.S. of the equation (ii) we get,

$$\begin{gathered} L.H.S.=2\left(2\right)-3\left(0\right)-5(-1) \\ =4-0+5 \\ =9 \\ \ne R.H.S \end{gathered}$$

Hence, option (d) is also incorrect.

Therefore, option (c) is the correct answer.