The equation of the plane containing the line\[\overrightarrow{r}=\widehat{i}+\widehat{j}+t(2\widehat{i}+\widehat{j}+4\widehat{k})\], is
A \[\overrightarrow{r}.(\widehat{i}+2\widehat{j}-\widehat{k})=3\]
B \[\overrightarrow{r}.(\widehat{i}+2\widehat{j}-\widehat{k})=6\]
C \[\overrightarrow{r}.(-\widehat{i}-2\widehat{j}-\widehat{k})=3\]
D None of these

Hint: In the given options, the equation of the plane is given [r (xi + yj + zc) = d], where r is the position vector of the arbitrary point on the line which passes through the plane] but we need to find the equation of the plane which contains the line \[\overrightarrow{r}=\widehat{i}+\widehat{j}+t(2\widehat{i}+\widehat{j}+4\widehat{k})\]or can say the plane to which the given line satisfy its equation. For this, we will check all planes with the given value of r and will check if the value on the left-hand side is equal to the right-hand side or not.

Complete step by step solution: In this question, we need to check all the equation of the plane given in the option by putting the value of r which is given in question (equation of line) or we can say we need to find the equation of the plane which satisfy the given equation of line such as
\[\overrightarrow{r}=\widehat{i}+\widehat{j}+t(2\widehat{i}+\widehat{j}+4\widehat{k})\]
Modifying it we get
\[\overrightarrow{r}=(1+2t)\widehat{i}+(1+t)\widehat{j}+4t\widehat{k}\]
Now as we put the above value of r in option C such as
\[\overrightarrow{r}.(-\widehat{i}-2\widehat{j}-\widehat{k})=3\] (Plane equation given in option C)
\[\{(1+2t)\widehat{i}+(1+t)\widehat{j}+4t\widehat{k}\}(-\widehat{i}-2\widehat{j}-\widehat{k})\]
\[\left( 1+2t \right)\left( -1 \right)\text{ }+\text{ }\left( 1+t \right)\left( -2 \right)\text{ }+\text{ }4t\left( -1 \right)\]
     \[-1\text{ }\text{ }2t\text{ }\text{ }2\text{ }\text{ }2t\text{ }\text{ }4t\] Is not equal to 3 so the given line equation does not satisfy this plane so the plane in option C is not the plane passing through the given line.

Now as we put the above value of r in option B such as
\[\overrightarrow{r}.(\widehat{i}+2\widehat{j}-\widehat{k})=6\] (Plane equation given in option B)
\[\{(1+2t)\widehat{i}+(1+t)\widehat{j}+4t\widehat{k}\}(\widehat{i}+2\widehat{j}-\widehat{k})\]
\[\left( 1+2t \right)\left( 1 \right)\text{ }+\text{ }\left( 1+t \right)\left( 2 \right)\text{ }+\text{ }4t\left( -1 \right)\]
     \[1\text{ }+\text{ }2t\text{ }+\text{ }2\text{ }+\text{ }2t\text{ }\text{ }4t\]Is not equal to 6 so the given line equation does not satisfy this plane so the plane in option B is not the plane passing through the given line.

Now as we put the above value of r in option A such as
\[\overrightarrow{r}.(\widehat{i}+2\widehat{j}-\widehat{k})=3\] (Plane equation given in option A)
\[\{(1+2t)\widehat{i}+(1+t)\widehat{j}+4t\widehat{k}\}(\widehat{i}+2\widehat{j}-\widehat{k})\]
\[\left( 1+2t \right)\left( 1 \right)\text{ }+\text{ }\left( 1+t \right)\left( 2 \right)\text{ }+\text{ }4\left( -1 \right)\]
\[1\text{ }+\text{ }2t\text{ }+\text{ }2\text{ }+\text{ }2t\text{ }\text{ }4t\] Is equal to 3 so the given line equation does satisfy this plane so the plane in option A is the plane passing through the given line.

Option ‘A’ is correct

Note: It is important to note that the dot product of the same unit vector is equal to one or can say in the dot product coefficient of the I unit vector will get multiplied by the coefficient of the I unit vector such as
\[\left( x\widehat{i} \right)\left( a\widehat{i} \right)\text{ }=\text{ }xa\]
The dot product between different unit vectors is equal to zero such as
\[\left( x\widehat{i} \right)\left( b\widehat{j} \right)\text{ }=\text{ }0\]
But in cross product opposite unit vectors interact with each other and the cross product of the same unit vector is zero.