Find the equation of the plane passing through

Any plane parallel to the plane, $\vec{r}_{1} \cdot(\hat{i}+\hat{j}+\hat{k})=2$, is of the form $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=\lambda$        ...(1)

The plane passes through the point $(a, b, c)$. Therefore, the position vector $\vec{r}$ of this point is $\vec{r}=a \hat{i}+b \hat{j}+c \hat{k}$

Therefore, equation (1) becomes

$(a \hat{i}+b \hat{j}+c \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=\lambda$

$\Rightarrow a+b+c=\lambda$

Substituting $\lambda=a+b+c$ in equation (1), we obtain

$\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=a+b+c$            ...(2)

This is the vector equation of the required plane.

Substituting $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$ in equation (2), we obtain

$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=a+b+c$

$\Rightarrow x+y+z=a+b+c$