The vector equation of the plane through the line of intersection of the planes $x+y+z=1$ and \[2x+3y+4z=5\] which is perpendicular to the plane $x-y+z=0$ is:
A. $\vec{r}\times \left( \hat{i}+\hat{k} \right)+2=0$
B. $\vec{r}.\left( \hat{i}-\hat{k} \right)-2=0$
C. $\vec{r}.\left( \hat{i}-\hat{k} \right)+2=0$
D. $\vec{r}\times \left( \hat{i}-\hat{k} \right)+2=0$

Hint: We will first find the equation of a plane passing through the intersection of planes using the conventional formula, it will have $\lambda $ as unknown. To find this value we will use the second part of the question and will use the condition ${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$ for the normal vectors of both the planes. Finally we will convert the scalar equation into its vector form.

Complete step by step answer:
We know that Equation of a plane passing through the intersection of planes ${{A}_{1}}x+{{B}_{1}}y+{{C}_{1}}z={{d}_{1}}$and ${{A}_{2}}x+{{B}_{2}}y+{{C}_{2}}z={{d}_{2}}$ is $\left( {{A}_{1}}x+{{B}_{1}}y+{{C}_{1}}z-{{d}_{1}} \right)+\lambda \left( {{A}_{2}}x+{{B}_{2}}y+{{C}_{2}}z-{{d}_{2}} \right)=0\text{ }........\text{Equation 1}\text{.}$
Now we have to take our first plane equation: $x+y+z=1$ and compare it with our standard equation written above: ${{A}_{1}}x+{{B}_{1}}y+{{C}_{1}}z={{d}_{1}}$ , to find out the values of ${{A}_{1}},{{B}_{1}},{{C}_{1}}$ and ${{d}_{1}}$.
Now the plane equation can be written as follows,
$x+y+z=1\Rightarrow 1x+1y+1z=1$ , on comparing it with ${{A}_{1}}x+{{B}_{1}}y+{{C}_{1}}z={{d}_{1}}$ we will get the following values:
${{A}_{1}}=1,{{B}_{1}}=1,{{C}_{1}}=1,{{d}_{1}}=1\text{ }.............\text{ Equation 2}\text{.}$
Similarly for the second equation of plane $2x+3y+4z=5$ and compare it with our standard equation written above: ${{A}_{2}}x+{{B}_{2}}y+{{C}_{2}}z={{d}_{2}}$ , to find out the values of ${{A}_{2}},{{B}_{2}},{{C}_{2}}$ and ${{d}_{2}}$.
Now on comparing it with ${{A}_{2}}x+{{B}_{2}}y+{{C}_{2}}z={{d}_{2}}$ we will get the following values:
${{A}_{2}}=2,{{B}_{2}}=3,{{C}_{2}}=4,{{d}_{2}}=5.............\text{ Equation 3}\text{.}$
Now we have written the formula of equation of plane through the intersection of the plane in equation 1 as : $\left( {{A}_{1}}x+{{B}_{1}}y+{{C}_{1}}z-{{d}_{1}} \right)+\lambda \left( {{A}_{2}}x+{{B}_{2}}y+{{C}_{2}}z-{{d}_{2}} \right)=0$ . Now we will put the values from equation 2: ${{A}_{1}}=1,{{B}_{1}}=1,{{C}_{1}}=1,{{d}_{1}}=1$ and equation 3: ${{A}_{2}}=2,{{B}_{2}}=3,{{C}_{2}}=4,{{d}_{2}}=5$ in this equation:
$\begin{align}
  & \Rightarrow (1x+1y+1z-1)+\lambda \left( 2x+3y+4z-5 \right)=0 \\
 & \Rightarrow x+y+z-1+2\lambda x+3\lambda y+4\lambda z-5\lambda =0 \\
\end{align}$
Now, we will take the variables x, y, z out:
$\Rightarrow \left( 1+2\lambda \right)x+\left( 1+3\lambda \right)y+\left( 1+4\lambda \right)z+\left( -1-5\lambda \right)=0\text{ }................\text{Equation 4}\text{.}$

Now as given in the question, the plane is also perpendicular to the plane $x-y+z=0$ . So, the normal vector $\vec{N}$ of the required plane will be perpendicular to the normal vector $\vec{n}$ of $x-y+z=0$.
We will now find $\vec{N}$ and $\vec{n}$ :
We have the required plane from Equation 4: $\left( 1+2\lambda \right)x+\left( 1+3\lambda \right)y+\left( 1+4\lambda \right)z+\left( -1-5\lambda \right)=0$ , Therefore, $\vec{N}=\left( 1+2\lambda \right)\hat{i}+\left( 1+3\lambda \right)\hat{j}+\left( 1+4\lambda \right)\hat{k}$
Similarly, for the plane $x-y+z=0$ , \[\vec{n}=1\hat{i}-1\hat{j}+1\hat{k}\].
Now we know that: Two lines with direction ratios ${{a}_{1}},{{b}_{1}},{{c}_{1}}$ and ${{a}_{2}},{{b}_{2}},{{c}_{2}}$ are perpendicular, if ${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0\Rightarrow \text{ Theory 1}\text{.}$
We will find the direction ratios of $\vec{N}$ and $\vec{n}$:
For $\vec{N}=\left( 1+2\lambda \right)\hat{i}+\left( 1+3\lambda \right)\hat{j}+\left( 1+4\lambda \right)\hat{k}$ : Direction Ratios will be $1+2\lambda ,1+3\lambda ,1+4\lambda $ $\begin{align}
  & {{a}_{1}}=1+2\lambda \\
 & {{b}_{1}}=1+3\lambda \\
 & {{c}_{1}}=1+4\lambda \\
\end{align}$
Similarly for \[\vec{n}=1\hat{i}-1\hat{j}+1\hat{k}\], Direction ratios will be $1,-1,1$ ; Therefore
$\begin{align}
  & {{a}_{2}}=1 \\
 & {{b}_{2}}=-1 \\
 & {{c}_{2}}=1 \\
\end{align}$

Now since $\vec{N}$ and $\vec{n}$ are perpendicular: ${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$ , as we saw in Theory 1, now we will put the values of direction ratios in this equation:
\[\begin{align}
  & \Rightarrow {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0 \\
 & \Rightarrow \left( \left( 1+2\lambda \right)\times 1 \right)+\left( \left( 1+3\lambda \right)\times \left( -1 \right) \right)+\left( \left( 1+4\lambda \right)\times 1 \right)=0 \\
 & \Rightarrow 1+2\lambda -1-3\lambda +1+4\lambda =0 \\
 & \Rightarrow 1+3\lambda =0 \\
 & \Rightarrow \lambda =\dfrac{-1}{3} \\
\end{align}\]
Putting Value of $\lambda $ in equation 4, we have:
$\begin{align}
  & \left( 1+2\lambda \right)x+\left( 1+3\lambda \right)y+\left( 1+4\lambda \right)z+\left( -1-5\lambda \right)=0 \\
 & \Rightarrow \left( 1+2\times \left( \dfrac{-1}{3} \right) \right)x+\left( 1+3\times \left( \dfrac{-1}{3} \right) \right)y+\left( 1+4\times \left( \dfrac{-1}{3} \right) \right)z+\left( -1-5\times \left( \dfrac{-1}{3} \right) \right)=0 \\
 & \Rightarrow \left( 1-\dfrac{2}{3} \right)x+\left( 1-1 \right)y+\left( 1-\dfrac{4}{3} \right)z+\left( -1+\dfrac{5}{3} \right)=0 \\
 & \Rightarrow \dfrac{1}{3}x-\dfrac{1}{3}z+\dfrac{2}{3}=0 \\
 & \Rightarrow \dfrac{1}{3}\left( x-z+2 \right)=0 \\
 & \Rightarrow x-z+2=0 \\
\end{align}$
We have with us the equation of plane as : $x-z+2=0$ , now we will convert it into the vector form:
Let $\overrightarrow{r}$ be the direction vector and therefore the vector equation becomes: $\overrightarrow{r}\left( \hat{i}-\hat{k} \right)+2=0$
Hence the answer is Option C.

Note:
Student might get confused in putting the values of the ${{A}_{1}},{{B}_{1}},{{C}_{1}}$ and ${{d}_{1}}$ and similarly other values as well. Chances of silly mistakes are more here as the values are very small and are used frequently. After finding the value of $\lambda $ , take care of the negative sign in the calculation.