Find the equation of plane that contains the point A(2. 1, -1) and is perpendicular to the line of intersection of the planes 2x+y-z=3 and x+2y+z=2. Also find the angle between the plane thus obtained and the y-axis.

OR

Find the distance of the point P(-2, -4, 7)from the point of intersection Q of the line $\overrightarrow{\mathrm{r}}=$$(3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+6\hat{\mathrm{k}})+\mathrm{\lambda }(2\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\mathrm{k})$ and the plane $\overrightarrow{\mathrm{r}}\cdot (\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})=6$ Also write the vector equation of the line PQ.

Given:
Equation of required plane contain the point A(2, 1, -1),
Hence, by using the standard equation of the plane a(x-x1)+b(y-y1)+c(z-z1)=0, where (x1, y1, z1)=(2, 1,-1). It follows,
a(x-2)+b(y-1)+c(z+1)=0
From the question given that plane is perpendicular to the line of intersection of 2x+y−z=3 and
x+2y+z=2,
Hence, it’s coefficient dot product will zero,
$\therefore (2,1,-1)\cdot (\mathrm{a},\mathrm{b},\mathrm{c})=0\Rightarrow 2\mathrm{a}+\mathrm{b}-\mathrm{c}=0...\left(1\right) (\mathrm{For} \mathrm{line} 2\mathrm{x}+\mathrm{y}-\mathrm{z}=3)$
For line x+2y+z=2,
(1, 2, 1) ⋅(a,b,c) = 0 $\Rightarrow \mathrm{a}+2\mathrm{b}+\mathrm{c}=0....\left(2\right)$
Let add the equation (1) in (2),
$$\begin{gathered} \therefore (2\mathrm{a}+\mathrm{b}-\mathrm{c})+(\mathrm{a}+2\mathrm{b}+\mathrm{c})=0+0 \\ \therefore 3\mathrm{a}+3\mathrm{b}=0\Rightarrow \mathrm{a}-\mathrm{b} \end{gathered}$$
Substitute the value of a=-b in equation (1),
$\begin{array}{l}\therefore 2\mathrm{a}-\mathrm{a}-\mathrm{c}=0\Rightarrow \mathrm{a}=\mathrm{c}\\ \therefore \mathrm{a}=-\mathrm{b}=\mathrm{c}=\mathrm{k}\text{ (Assume) }\end{array}$
Now, the equation of the plane,
$\mathrm{a}(\mathrm{x}-2)+\mathrm{b}(\mathrm{y}-1)+\mathrm{c}(\mathrm{z}+1)=0$
Substitute $\mathrm{a}=-\mathrm{b}=\mathrm{c}=\mathrm{k}$ in the equation.
$\begin{array}{l}\therefore \mathrm{k}(\mathrm{x}-2)-\mathrm{k}(\mathrm{y}-1)+\mathrm{k}(\mathrm{z}+1)=0\\ \therefore \mathrm{x}-2-\mathrm{y}+1+\mathrm{z}+1=0\\ \therefore \mathrm{x}-\mathrm{y}+\mathrm{z}=0\end{array}$
Now, recall the formula for angle between line and plane which is given by,
$\sin \mathrm{\theta }=\frac{\mathrm{la}+\mathrm{mb}+\mathrm{nc}}{\sqrt{{\mathrm{l}}^2+{\mathrm{m}}^2+{\mathrm{n}}^2}\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}$
Now, the angle between y-axis(D.R’s (0, 1, 0)) and x-y+z=0 is,
Here, (a, b, c)=(0, 1, 0) and (l, m, n)=(1, -1, 1)
$\begin{array}{l}\therefore \sin \mathrm{\theta }=\left|\frac{\left(1\right)0+\left(1\right)(-1)+\left(1\right)0}{\sqrt{1^2+(-1{)}^2+1^2}\sqrt{0^2+1^2+0^2}}\right|\\ \therefore \sin \mathrm{\theta }=\left|\frac{0-1+0}{\sqrt{1+1+1}\sqrt{1}}\right|\\ \therefore \sin \mathrm{\theta }=\left|\frac{-1}{\sqrt{3}}\right|\\ \therefore \sin \mathrm{\theta }=\frac{1}{\sqrt{3}}\\ \therefore \mathrm{\theta }={\sin }^{-1} \left(\frac{1}{\sqrt{3}}\right)\end{array}$
Therefore, the equation of plane is $\mathrm{x}-\mathrm{y}+\mathrm{z}=0$ and the angle between y-axis and the plane is $\mathrm{\theta }=$${\sin }^{-1} \left(\frac{1}{\sqrt{3}}\right)$

OR

Given,

The line of equation:$\overrightarrow{\mathrm{r}}=(3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+6\hat{\mathrm{k}})+\mathrm{\lambda }(2\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\widehat{\mathrm{kj}})$
And the plane of the equation: $\overrightarrow{\mathrm{r}}\cdot (\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})=6$
To find the point of intersection of line and plane, we will putting the value of$\overrightarrow{\mathrm{r}}$from the equation of line into the equation of plane,
$\begin{array}{l}\therefore \left[\right(3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+6\hat{\mathrm{k}})+\mathrm{\lambda }(2\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}]\cdot (\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}=6\\ \therefore [3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+6\hat{\mathrm{k}}+2\mathrm{\lambda }\hat{\mathrm{i}}-\mathrm{\lambda }\hat{\mathrm{j}}+2\mathrm{\lambda }\hat{\mathrm{k}}]\cdot (1\hat{\mathrm{i}}-1\hat{\mathrm{j}}+1\hat{\mathrm{k}})=6\\ \therefore \left[\right(3+2\mathrm{\lambda })\hat{\mathrm{i}}+(-2-\mathrm{\lambda })\hat{\mathrm{j}}+(6+2\mathrm{\lambda }\left)\hat{\mathrm{k}}\right]\cdot (1\hat{\mathrm{i}}-1\hat{\mathrm{j}}+1\hat{\mathrm{k}})=6\\ \therefore (3+2\mathrm{\lambda })\times 1+(-2-\mathrm{\lambda })\times (-1)+(6+2\mathrm{\lambda })\times 1=6\\ \therefore 3+2\mathrm{\lambda }+2+\mathrm{\lambda }+6+2\mathrm{\lambda }=6\\ \therefore 5\mathrm{\lambda }+11=6\\ \therefore 5\mathrm{\lambda }=6-11\\ \therefore 5\mathrm{\lambda }=-5\\ \therefore \mathrm{\lambda }=-1\end{array}$
Substitute this value in the equation of line,
$\begin{array}{l}\therefore \overrightarrow{\mathrm{r}}=(3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+6\hat{\mathrm{k}})+(-1)(2\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}})\\ \therefore \overrightarrow{\mathrm{r}}=3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+6\hat{\mathrm{k}}-2\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}}\\ \therefore \overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+4\hat{\mathrm{k}}\end{array}$
Let the point of intersection be (x, y, z),
Hence, $\overrightarrow{\mathrm{r}}=\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}+\mathrm{z}\hat{\mathrm{k}}$
$\hat{\mathrm{i}}-\hat{\mathrm{j}}+4\hat{\mathrm{k}}=\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}+\mathrm{z}\hat{\mathrm{k}}$
Therefore, x=1, y=-1 and z=4.
So the point of intersection is Q(1, -1, 4).
Now, we know that the distance between two points (x1, y1, z1) and (x2, y2, z2) is
$\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2+{\left({\mathrm{z}}_2-{\mathrm{z}}_1\right)}^2}$
For distance between P(-2, -4, 7) and Q(1, -1, 4),
$\begin{array}{l}=\sqrt{(1-(-2){)}^2+((-1)-(-4){)}^2+(4-7{)}^2}\\ =\sqrt{3^2+3^2+(-3{)}^2}\end{array}$
$=\sqrt{9+9+9}=\sqrt{27}=3\sqrt{3}$
For vector equation of PQ,
Given that PQ passes through P(-2, -4, 7), so its position vector is given by,
$\overrightarrow{\mathrm{a}}=-2\hat{\mathrm{i}}-4\hat{\mathrm{j}}+7\hat{\mathrm{k}}$
Now, for direction ration of PQ,
$(1-(-2\left)\right)=3,(-1-(-4\left)\right)=3\text{ and }(4-7)=-3$
So, the equation of the vector in the direction of PQ is,
$\overrightarrow{\mathrm{b}}=3\hat{\mathrm{i}}+3\hat{\mathrm{j}}-3\hat{\mathrm{k}}$
Recall the equation of vector form $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\mathrm{\mu }\overrightarrow{\mathrm{b}},\mathrm{\mu }\in \mathrm{R}$ For PQ
$\therefore \overrightarrow{\mathrm{r}}=-2\hat{\mathrm{i}}-4\hat{\mathrm{j}}+7\hat{\mathrm{k}}+\mathrm{\mu }(3\hat{\mathrm{i}}+3\hat{\mathrm{j}}-3\hat{\mathrm{k}})$
Therefore, distance between the points PQ is $3\sqrt{3}$ and the vector equation of the line PQ is$\overrightarrow{\mathrm{r}}=$$-2\hat{\mathrm{i}}-4\hat{\mathrm{j}}+7\hat{\mathrm{k}}+\mathrm{\mu }(3\hat{\mathrm{i}}+3\hat{\mathrm{j}}-3\hat{\mathrm{k}})$.