A plane pi passing through the point 3 hat{i} | vedclass

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(C) The equation of a plane passing through a point $\vec{a}$ and perpendicular to a normal vector $\vec{n}$ is given by $(\vec{r}-\vec{a}) \cdot \vec

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$x+2y-3z+20=0$

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(C) The equation of a plane passing through a point $\vec{a}$ and perpendicular to a normal vector $\vec{n}$ is given by $(\vec{r}-\vec{a}) \cdot \vec{n} = 0$. First, find the equation of the plane passing through $\hat{i}+\hat{j}-\hat{k}$ with normal vector $\vec{n} = \hat{i}+2\hat{j}-3\hat{k}$: $1(x-1) + 2(y-1) - 3(z+1) = 0$ $x - 1 + 2y - 2 - 3z - 3 = 0$ $x + 2y - 3z - 6 = 0$. Since plane $\pi$ is parallel to this plane, it will have the same normal vector $\vec{n} = \hat{i}+2\hat{j}-3\hat{k}$. Thus, the equation of plane $\pi$ passing through $3\hat{i}-4\hat{j}+5\hat{k}$ is: $1(x-3) + 2(y+4) - 3(z-5) = 0$ $x - 3 + 2y + 8 - 3z + 15 = 0$ $x + 2y - 3z + 20 = 0$.

$A$ plane $\pi$ passing through the point $3 \hat{i}-4 \hat{j}+5 \hat{k}$ is parallel to the plane which passes through the point $\hat{i}+\hat{j}-\hat{k}$ and is perpendicular to the vector $\hat{i}+2 \hat{j}-3 \hat{k}$. Then the Cartesian equation of $\pi$ is

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