# 8. Find the equation of the plane passing through (a, b, c) and parallel to the plane $\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2$.

Given that the plane is passing through $(a,b,c)$ and is parallel to the plane $\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2$

So, we have

The position vector of the point $A(a,b,c)$ is, $\vec{r_{A}} = a\widehat{i}+b\widehat{j}+c\widehat{k}$

and any plane which is parallel to the plane, $\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2$ is of the form,

$\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=\lambda$.                      .......................(1)

Therefore the equation we get,

$( a\widehat{i}+b\widehat{j}+c\widehat{k}).(\widehat{i}+\widehat{j}+\widehat{k})=\lambda$

Or, $a+b+c = \lambda$

So, now substituting the value of $\lambda = a+b+c$ in equation (1), we get

$\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=a+b+c$            .................(2)

So, this is the required equation of the plane .

Now, substituting $\vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}$ in equation (2), we get

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

Or, $x+y+z = a+b+c$

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