A plane containing the point \[(3,2,0)\] and the line \[\dfrac{{{\text{x - 1}}}}{{\text{1}}}{\text{ = }}\dfrac{{{\text{y - 2}}}}{{\text{5}}}{\text{ = }}\dfrac{{{\text{z - 3}}}}{{\text{4}}}\] also contains the point
A) \[{\text{(0,7, - 10)}}\]
B) \[{\text{(0,7,10)}}\]
C) \[{\text{(0,3,1)}}\]
D) \[{\text{(0, - 3,1)}}\]

Hint: Firstly calculate the direction ratios of line from the points \[(3,2,0)\]and \[(1,2,3)\]. Now, as we are having the direction ratio of line and direction ratios calculated from the above points we can calculate the direction ratios of normal to the plane by taking cross products of both the direction ratios. And hence once its calculated than we can write the equation of plane as \[{\text{Ax + By + Cz = k}}\], here A, B, C are direction ratios and by putting any point lying in the plane we can calculate the value of k. Hence, once the equation of the plane is obtained by substituting the points given in the options, the point which satisfies the equation is our required answer.

Complete step by step solution: As the given points and lines are \[(3,2,0)\]and \[\dfrac{{{\text{x - 1}}}}{{\text{1}}}{\text{ = }}\dfrac{{{\text{y - 2}}}}{{\text{5}}}{\text{ = }}\dfrac{{{\text{z - 3}}}}{{\text{4}}}\].
Diagram:

As points \[(3,2,0)\]and \[(1,2,3)\]lies on the plane so the direction ratios of the line can be given as
\[
   \Rightarrow {\text{(3 - 1,2 - 2,0 - 3)}} \\
  {\text{ = (2,0, - 3)}} \\
 \]
And direction ratios of given line is \[(1,5,4)\].
Now, by taking cross product of both the above quantities we can direction ratios of normal to the plane as,
\[ \Rightarrow {\text{(}}\widehat {\text{i}}{\text{ + 5}}\widehat {\text{j}}{\text{ + 4}}\widehat {\text{k}}{\text{)}} \times {\text{(2}}\widehat {\text{i}}{\text{ - 3}}\widehat {\text{k}}{\text{)}}\]
Solving the above cross product by determinant method as,
\[
  {\text{ = }}\left| {\begin{array}{*{20}{c}}
  {\widehat {\text{i}}}&{\widehat {\text{j}}}&{\widehat {\text{k}}} \\
  {\text{1}}&{\text{5}}&{\text{4}} \\
  {\text{2}}&{\text{0}}&{{\text{( - 3)}}}
\end{array}} \right| \\
  {\text{ = - 15}}\widehat {\text{i}}{\text{ + 11}}\widehat {\text{j}}{\text{ - 10}}\widehat {\text{k}} \\
 \]
Hence, the equation of plane can be given as,
\[{\text{ - 15x + 11y - 10z = k}}\]
Satisfy any of the points which lie on the plane. Let the point be \[(1,2,3)\].
\[
  {\text{ - 15(1) + 11(2) - 10(3) = k}} \\
   \Rightarrow {\text{k = - 15 + 22 - 30}} \\
   \Rightarrow {\text{k = - 23}} \\
 \]
Now, putting the value of k in the above equation,
The equation of plane is finally given as \[{\text{ - 15x + 11y - 10z = - 23}}\]
Now, put the points one by one in the equation of the plane and the one which satisfies it will be our required answer.
Let us put first, \[{\text{(0,7, - 10)}}\]
On substituting the value of the point in \[{\text{ - 15x + 11y - 10z = - 23}}\], we get,
\[
  {\text{ - 15(0) + 11(7) - 10( - 10) = - 23}} \\
   \Rightarrow 77 + 100 = - 23 \\
   \Rightarrow 177 \ne - 23 \\
 \]
Now, check for option (B) \[{\text{(0,7,10)}}\],
On substituting the value of the point in \[{\text{ - 15x + 11y - 10z = - 23}}\], we get,
\[
  {\text{ - 15(0) + 11(7) - 10(10) = - 23}} \\
   \Rightarrow 77 - 100 = - 23 \\
   \Rightarrow - 23 = - 23 \\
 \]
As \[{\text{(0,7,10)}}\], satisfies the equation \[{\text{ - 15x + 11y - 10z = - 23}}\].

Hence, option (B) is our correct answer.

Note: A plane is a flat, two-dimensional surface that extends infinitely far. A plane is the two-dimensional analogue of a point, a line and three-dimensional space.
Calculate the cross product using the proper concept of \[ {A \times B =} \left| {\begin{array}{*{20}{c}}
  {\widehat {\text{i}}}&{\widehat {\text{j}}}&{\widehat {\text{k}}} \\
  {{{\text{x}}_{\text{1}}}}&{{{\text{y}}_{\text{1}}}}&{{{\text{z}}_{\text{1}}}} \\
  {{{\text{x}}_{\text{2}}}}&{{{\text{y}}_{\text{2}}}}&{{{\text{z}}_{\text{2}}}}
\end{array}} \right|\]. After calculating direction ratios of normal to the plane use it carefully in the equation of plane and form the equation of required plane.