# The plane passing through the points (1,2,1), (2,1,2) and

Question:

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

1. $(0,6,-2)$

2. $(-2,0,1)$

3. $(0,-6,2)$

4. $(2,0,-1)$

Correct Option: , 2

Solution:

Two points on the line ( $\mathrm{L}$ say) $\frac{\mathrm{x}}{3}=\frac{\mathrm{y}}{2}, \mathrm{z}=1$ are

$(0,0,1) \&(3,2,1)$

So dr's of the line is $\langle 3,2,0\rangle$

Line passing through $(1,2,1)$, parallel to $\mathrm{L}$ and coplanar with given plane is

$\overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}+\mathrm{t}(3 \hat{\mathrm{i}}+2 \mathrm{j}), \mathrm{t} \in \mathrm{R}(-2,0,1)$ satisfies

$\Rightarrow(-2,0,1)$ lies on given plane.

Answer of the question is $(2)$

We can check other options by finding eqution of plane

Equation plane : $\left|\begin{array}{lll}x-1 & y-2 & z-1 \\ 1+2 & 2-0 & 1-1 \\ 2+2 & 1-0 & 2-1\end{array}\right|=0$

$\Rightarrow 2(x-1)-3(y-2)-5(z-1)=0$

$\Rightarrow 2 x-3 y-5 z+9=0$