\[a = i + j + k,{\bf{b}} = 2i - 4k,c = i + \lambda j + 3k\] Are coplanar, then the value of \[\lambda \] is
A. \[\dfrac{5}{2}\]
B. \[\dfrac{3}{5}\]
C. \[\dfrac{7}{3}\]
D. None of these

Hint: To solve this problem, we apply the scalar product formula to these vectors. \[\vec a,\vec b,\vec c\] is the formula for the scalar product of three vectors \[a,b,c\]. The vector \[\left( {\overrightarrow b \times \overrightarrow c } \right)\] represents the vector product of two vectors. We use this information to solve the problem because the scalar product for coplanar vectors equals\[0\].



Formula Used:The scalars x, y, and z must not all be zero in order for the vectors a, b, and c to be coplanar.
\[{\bf{xa}} + {\bf{yb}} + {\bf{zc}} = 0\]



Complete step by step solution:We have been three vectors in the equation that, the vectors\[a = i + j + k,{\bf{b}} = 2i - 4k,c = i + \lambda j + 3k\] are coplanars
We already know that, three vectors, a, b, c are coplanar if
\[[{\rm{abc}}] = 0\]
For calculating the vector product of these vectors, we shall use matrix notation.
Now, we have to write the vectors in the form of matrix to determine the value of\[\lambda \], we obtain
OR \[\left| {\begin{array}{*{20}{c}}2&{ - 1}&1\\1&2&{ - 3}\\3&\lambda &5\end{array}} \right| = 0\]
On expanding along first row of the above matric we get,
\[2(10 + 3\lambda ) + 1(5 + 9) + 1(\lambda - 6) = 0\]
From the above equation, it is found that
\[\begin{array}{l}1 \cdot (5 + 9) = 14\\1 \cdot (\lambda - 6) = \lambda - 6\end{array}\]
Now, the above equation can be written as,
\[2\left( {10 + 3\lambda } \right) + 14 + \lambda - 6 = 0\]
Now, we have to subtract \[\left( {14 - 6} \right)\] from both sides of the above equation, we obtain
\[2\left( {10 + 3\lambda } \right) + 14 + \lambda - 6 - \left( {14 - 6} \right) = 0 - \left( {14 - 6} \right)\]
On simplifying the obtained equation, we get
\[2\left( {10 + 3\lambda } \right) + \lambda = - 8\]
Now, we have to expand the term \[2\left( {10 + 3\lambda } \right)\] in the above equation:
\[{\rm{20 + 6}}\lambda {\rm{ + }}\lambda {\rm{ = - 8}}\]
Now, let’s add similar terms, we obtain
\[{\rm{20 + 7}}\lambda {\rm{ = - 8}}\]
Solve the above equation by having constants on one side and variables on other side, we get
\[{\rm{7}}\lambda {\rm{ = - 8 - 20}}\]
Now, we have to simplify, we obtain
\[7\lambda + 28 = 0\]
Let us solve for \[\lambda \], to obtain the desired result:
\[\lambda = - 4\]
Therefore, the value of \[\lambda \] is \[\lambda = - 4\]



Option ‘D’ is correct

Note: The cyclic property of triple scalar can also be used to determine the requirement of coplanarity. As a result of this attribute, product discovering scalar triple product between vectors \[\vec b,\vec c,\vec a\]or \[\vec c,\vec a,\vec b\]instead of \[\vec a,\vec b,\vec c\]. As a result, we will get the same response in each of these other ways.