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The distance of the point $\left( {1,0,2} \right)$ from the point of intersection of the $\dfrac{{x - 2}}{3} = \dfrac{{y + 1}}{4} = \dfrac{{z - 2}}{{12}}$ and the plane $x - y + z = 16$ is.
A. $2\sqrt {14} $
B. $8$
C. $3\sqrt {21} $
D. $13$

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Last updated date: 04th Mar 2024
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Hint: First consider the given point, line and the plane. Next, let the equation of the line be equal to $\lambda $. Now, we will find the coordinates of the equation of line in terms of $\lambda $. After this we will put the coordinates in the equation of plane and determine the value of $\lambda $. Thus, we will get the coordinates by substituting the value of $\lambda $. As we need to find the distance so, we will apply the distance formula between the given point and the obtained point using $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $.

Complete step by step solution:
We will first consider the given: equation of line $\dfrac{{x - 2}}{3} = \dfrac{{y + 1}}{4} = \dfrac{{z - 2}}{{12}}$ and the equation of plane $x - y + z = 16$.
We will first put the equation of line equal to $\lambda $ and convert the coordinates in terms of $\lambda $.
Let the equation of line be $\dfrac{{x - 2}}{3} = \dfrac{{y + 1}}{4} = \dfrac{{z - 2}}{{12}}{\text{ }} \to \left( 1 \right)$
And equation of plane be $x - y + z = 16{\text{ }} \to \left( 2 \right)$
Now, we will put $\dfrac{{x - 2}}{3} = \dfrac{{y + 1}}{4} = \dfrac{{z - 2}}{{12}} = \lambda $
Next, to determine the points of intersection, we will write the coordinates in $\lambda $,
Thus, we get,
\[ \Rightarrow \left( {x,y,z} \right) = \left( {3\lambda + 2,4\lambda - 1,12\lambda + 2} \right)\]
Now we will put these values of points in the equation of the plane.
\[ \Rightarrow \left( {3\lambda + 2) - (4\lambda - 1) + (12\lambda + 2} \right) = 16\]
Now, we will solve the obtained equation for $\lambda $ .
\[
   \Rightarrow \left( {3\lambda + 2) - (4\lambda - 1) + (12\lambda + 2} \right) = 16 \\
   \Rightarrow 3\lambda + 2 - 4\lambda + 1 + 12\lambda + 2 = 16 \\
   \Rightarrow 11\lambda + 5 = 16 \\
   \Rightarrow 11\lambda = 16 - 5 \\
   \Rightarrow 11\lambda = 11 \\
   \Rightarrow \lambda = 1 \\
 \]
Now we will put value of $\lambda $ in equation of line to get the coordinates,
Thus, we get,
\[
   \Rightarrow \left( {x,y,z} \right) = \left( {3\left( 1 \right) + 2,4\left( 1 \right) - 1,12\left( 1 \right) + 2} \right) \\
   \Rightarrow \left( {x,y,z} \right) = \left( {5,3,14} \right) \\
 \]
Now, this is our point of intersection.
Now, we will find the distance between two points that are $\left( {{x_1},{y_1},{z_1}} \right) = \left( {1,0,2} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right) = \left( {5,3,14} \right)$ by using the formula$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $.
$
   \Rightarrow d = \sqrt {{{\left( {5 - 1} \right)}^2} + {{\left( {3 - 0} \right)}^2} + {{\left( {14 - 2} \right)}^2}} \\
   \Rightarrow d = \sqrt {{{\left( 4 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( {12} \right)}^2}} \\
   \Rightarrow d = \sqrt {16 + 9 + 144} \\
   \Rightarrow d = \sqrt {169} \\
   \Rightarrow d = 13 \\
 $
Thus, we get the distance between two pints is 13.
Hence, the correct option is D.

Note: The equation of line need to be put equal to $\lambda $ which will further give us the another point to determine the distance using $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $. As the equation of line and equation of plane get intersected so we can substitute the coordinates in terms of $\lambda $ in the equation of plane to get the value of $\lambda $.