The line through \[{\bf{i}} + 3{\bf{j}} + 2{\bf{k}}\] and perpendicular to the lines \[{\bf{r}} = ({\bf{i}} + 2{\bf{j}} - {\bf{k}}) + \lambda (2{\bf{i}} + {\bf{j}} + {\bf{k}})\]\[{\bf{r}} = (2{\bf{i}} + 6{\bf{j}} + {\bf{k}}) + \mu ({\bf{i}} + 2{\bf{j}} + 3{\bf{k}})\] is
A. \[{\bf{r}} = ({\bf{i}} + 2{\bf{j}} - {\bf{k}}) + \lambda ( - {\bf{i}} + 5{\bf{j}} - 3{\bf{k}})\]
B. \[{\bf{r}} = {\bf{i}} + 3{\bf{j}} + 2{\bf{k}} + \lambda ({\bf{i}} - 5{\bf{j}} + 3{\bf{k}})\]
C. \[{\bf{r}} = {\bf{i}} + 3{\bf{j}} + 2{\bf{k}} + \lambda ({\bf{i}} + 5{\bf{j}} + 3{\bf{k}})\]
D. \[{\bf{r}} = {\bf{i}} + 3{\bf{j}} + 2{\bf{k}} + \lambda ( - {\bf{i}} + 5{\bf{j}} - 3{\bf{k}})\]

Hint: In our case, we have been given in the question that, the lines are \[{\bf{r}} = ({\bf{i}} + 2{\bf{j}} - {\bf{k}}) + \lambda (2{\bf{i}} + {\bf{j}} + {\bf{k}})\]\[{\bf{r}} = (2{\bf{i}} + 6{\bf{j}} + {\bf{k}}) + \mu ({\bf{i}} + 2{\bf{j}} + 3{\bf{k}})\]here, we have to determine the line through \[{\bf{i}} + 3{\bf{j}} + 2{\bf{k}}\] and perpendicular to the given lines. For that, we have to write the coefficients of the given vectors in matrix form and then determine the determinant to obtain the desired solution.

FORMULA USED:
For the matrix A,
\[A = \left( {\begin{array}{*{20}{c}}a&b&c\\d&e&f\\g&h&i\end{array}} \right)\]
The determinant can be calculated as,
\[\left| A \right| = a\left( {ei - fh} \right) - b\left( {di - fg} \right) + c\left( {dh - eg} \right)\]



Complete step by step solution:We have been given in the question that, the lines are
\[{\bf{r}} = ({\bf{i}} + 2{\bf{j}} - {\bf{k}}) + \lambda (2{\bf{i}} + {\bf{j}} + {\bf{k}})\]\[{\bf{r}} = (2{\bf{i}} + 6{\bf{j}} + {\bf{k}}) + \mu ({\bf{i}} + 2{\bf{j}} + 3{\bf{k}})\]
Here, we have to determine the line through \[{\bf{i}} + 3{\bf{j}} + 2{\bf{k}}\] and perpendicular to the given lines.
Because the line is perpendicular to both of the provided lines
Therefore, the needed line's direction ratio is
\[ = \left( {\begin{array}{*{20}{c}}i&j&k\\2&1&1\\1&2&3\end{array}} \right)\]
To calculate the determinant of a \[3 \times 3\] matrix, take the first element of the first row and multiply it by a secondary \[2 \times 2\] matrix composed of the other components in the \[3 \times 3\] matrix that do not belong to the row or column to which your first selected element belongs.
Now, we have to calculate the determinant of the above matrix, we get\[ = i(3 - 2) - j(6 - 1) + k(4 - 1)\]
Now, we have to solve the terms inside the parentheses, we get
\[ = i(1) - j(5) + k(3)\]
Let us write the line of equation as below,
\[{\rm{\vec r}} = {\rm{i}} + 3{\rm{j}} + 2{\rm{k}} + \lambda ({\rm{i}} - 5{\rm{j}} + 3{\rm{k}})\]
On restructuring the above expression, we have
\[{\rm{\vec r}} = {\rm{i}} + 3{\rm{j}} + 2{\rm{k}} + \lambda ( - {\rm{i + }}5{\rm{j}} - 3{\rm{k}})\]
Therefore, the line through \[{\bf{i}} + 3{\bf{j}} + 2{\bf{k}}\] and perpendicular to the lines \[{\bf{r}} = ({\bf{i}} + 2{\bf{j}} - {\bf{k}}) + \lambda (2{\bf{i}} + {\bf{j}} + {\bf{k}})\]\[{\bf{r}} = (2{\bf{i}} + 6{\bf{j}} + {\bf{k}}) + \mu ({\bf{i}} + 2{\bf{j}} + 3{\bf{k}})\] is \[{\rm{\vec r}} = {\rm{i}} + 3{\rm{j}} + 2{\rm{k}} + \lambda ( - {\rm{i + }}5{\rm{j}} - 3{\rm{k}})\]


Thus, Option (D) is correct.

Note:Student should keep in mind that, solving matrices and determinants should be done carefully. Otherwise, it will lead to greater confusion and also time consuming. So, one should always remember that adding every element in that row or column multiplied by its cofactor. The determinant is the outcome.