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# The equation of a plane passing through the line of intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance $2\sqrt 3$ from the point $\left( {3,{\text{ }}1,{\text{ }} - 1} \right)$ isA.$5x - 11y + z = 17$B.$\sqrt 2 x + y = 3\sqrt 2 - 1$C.$x + y + z = \sqrt 3$D.$x - \sqrt 2 y = 1 - \sqrt 2$

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Hint: In order to determine the required equation of a plane passing through the line intersection of the plane $x + 2y + 3z = 2$ and $x - y + z = 3$.First, we compare this equation with the two plane ${{\rm P}_1} + \lambda {{\rm P}_2} = 0,\lambda \in \mathbb{R}$ then we get the equation as $ax + by + cz + d = 0$ with the points
.The distance between the line is $2\sqrt 3$. Adding the two planes does not yield their line of intersection. In fact, it is the equation of a plane passing through their line of intersection, rather than a line with the point $\left( {3,{\text{ }}1,{\text{ }} - 1} \right)$ as $(x,y,z)$. We use the formula for the plane distance is $D = \dfrac{{\left| {ax + by + cz + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$. We need to solve the question to get the required solution.

In this given problem,
We are given the line of intersection of planes $x + 2y + 3z = 2$and $x - y + z = 3$ at a distance $2\sqrt 3$ from the point $\left( {3,{\text{ }}1,{\text{ }} - 1} \right)$.
In this question we are supposed to find out the equation of a plane passing through the line of intersection of the planes
Let us consider the two plane equation as${P_1}:x + 2y + 3z = 2$and ${P_2}:x - y + z = 3$,
Let $D$ be the distance,$D = \dfrac{2}{{\sqrt 3 }}$
For this we have to first determine the plane equation ${{\rm P}_1} + \lambda {{\rm P}_2} = 0$ as $(x + 2y + 3z - 2) + \lambda (x - y + z - 3) = 0$ to simplify the this equation as the form $ax + by + cz + d = 0$ with the point $(x,y,z)$ as $\left( {3,{\text{ }}1,{\text{ }} - 1} \right)$.
Now, we have to simplify the equation $(x + 2y + 3z - 2) + \lambda (x - y + z - 3) = 0$
$(x + 2y + 3z - 2 + \lambda x - \lambda y + \lambda z - 3\lambda ) = 0 \\ (1 + \lambda )x + (2 - \lambda )y + (3 + \lambda )z - (2 + 3\lambda ) = 0 \to (3) \\$
Here, directly compare the equation$(1 + \lambda )x + (2 - \lambda )y + (3 + \lambda )z - (2 + 3\lambda ) = 0$ with$ax + by + cz + d = 0$. Where, $a = (1 + \lambda ),b = (2 - \lambda ),c = (3 + \lambda ),d = (2 + 3\lambda )$
The formula for the plane distance is $D = \dfrac{{\left| {ax + by + cz + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$
We have the values $a = (1 + \lambda ),b = (2 - \lambda ),c = (3 + \lambda ),d = (2 + 3\lambda )$and the given distance $D = \dfrac{2}{{\sqrt 3 }}$ By substitute all the values and points $\left( {3,{\text{ }}1,{\text{ }} - 1} \right)$ into the equation to find out the value$\lambda$
$\Rightarrow \dfrac{2}{{\sqrt 3 }} = \dfrac{{\left| {3(1 + \lambda ) + (2 - \lambda ) - (3 + \lambda ) - (2 + 3\lambda )} \right|}}{{\sqrt {{{(1 + \lambda )}^2} + {{(2 - \lambda )}^2} + {{(3 + \lambda )}^2}} }}$.
Comparing the denominator${(1 + \lambda )^2},({(2 - \lambda )^2},{(3 + \lambda )^2}$ with the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab,{(a - b)^2} = {a^2} + {b^2} - 2ab$ by expanding the bracket on RHS, we get
$\Rightarrow \dfrac{2}{{\sqrt 3 }} = \dfrac{{\left| {3 + 3\lambda + 2 - \lambda - 3 - \lambda - 2 - 3\lambda } \right|}}{{\sqrt {1 + {\lambda ^2} + 2\lambda + 4 + {\lambda ^2} - 4\lambda + 9 + {\lambda ^2} + 6\lambda } }} \\ \Rightarrow \dfrac{2}{{\sqrt 3 }} = \dfrac{{\left| { - 2\lambda } \right|}}{{\sqrt {3{\lambda ^2} + 4\lambda + 14} }} \;$
Take square on both sides of numerator and denominator, we get
$\Rightarrow \dfrac{{{{(2)}^2}}}{{{{(\sqrt 3 )}^2}}} = \dfrac{{{{( - 2\lambda )}^2}}}{{{{(\sqrt {3{\lambda ^2} + 4\lambda + 14} )}^2}}} \\ \Rightarrow \dfrac{4}{3} = \dfrac{{4{\lambda ^2}}}{{(3{\lambda ^2} + 4\lambda + 14)}}$
By simplify in further step to solving the fraction on both sides, we get
$4(3{\lambda ^2} + 4\lambda + 14) = 4(3){\lambda ^2} \\ 3{\lambda ^2} + 4\lambda + 14 = 3{\lambda ^2} \;$
Combining all the terms together to find out the value,$\lambda$, we get
$3{\lambda ^2} + 4\lambda + 14 - 3{\lambda ^2} = 0 \Rightarrow 4\lambda + 14 = 0 \\ 4\lambda = - 14 \Rightarrow \lambda = - \dfrac{{14}}{4} \;$
Therefore, the value of $\lambda = - \dfrac{7}{2}$. Then
Let the equation $(1 + \lambda )x + (2 - \lambda )y + (3 + \lambda )z - (2 + 3\lambda ) = 0$ on comparing with $ax + by + cz + d = 0$ and substitute the $\lambda = - \dfrac{7}{2}$ into plane equation.
$\lambda = - \dfrac{7}{2} \Rightarrow \left( {1 - \dfrac{7}{2}} \right)x + \left( {2 + \dfrac{7}{2}} \right)y + \left( {3 - \dfrac{7}{2}} \right)z - \left( {2 + 3\left( {\dfrac{{ - 7}}{2}} \right)} \right) = 0 \;$
Take LCM on the above equation, so we get
$\left( {\dfrac{{2 - 7}}{2}} \right)x + \left( {\dfrac{{4 + 7}}{2}} \right)y + \left( {\dfrac{{6 - 7}}{2}} \right)z - \left( {\dfrac{{4 - 21}}{2}} \right) = 0 \\ - \dfrac{5}{2}x + \dfrac{{11}}{2}y - \dfrac{1}{2}z + \dfrac{{17}}{2} = 0 \;$
We perform multiplication on both sides by $2$ to simplify the fraction, we can get
$- 5x + 11y - z + 17 = 0$
Therefore, The equation of a plane passing through the line of intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance $2\sqrt 3$from the point $\left( {3,{\text{ }}1,{\text{ }} - 1} \right)$ is $- 5x + 11y - z + 17 = 0$
So, the final answer is option A: $5x - 11y + z = 17$
So, the correct answer is “Option A”.

Note: In this problem, always try to understand the mathematical statement .when using the distance formula, use the notation for the point $\left( {3,{\text{ }}1,{\text{ }} - 1} \right)$ as $(x,y,z)$. This will help to find out the final equation by substituting the values into the formula.