vec{n} is a unit vector normal to the plane p | vedclass

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(B) The equation of the plane $\pi$ containing two vectors $\vec{\alpha}$ and $\vec{\beta}$ and passing through point $\vec{a}$ is given by $(\vec{r}-

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$8$

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(B) The equation of the plane $\pi$ containing two vectors $\vec{\alpha}$ and $\vec{\beta}$ and passing through point $\vec{a}$ is given by $(\vec{r}-\vec{a}) \cdot (\vec{\alpha} 	imes \vec{\beta}) = 0$. First, calculate the normal vector $\vec{n}' = \vec{\alpha} 	imes \vec{\beta}$: $\vec{n}' = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 0 & 3 \ 2 & 1 & -1 \end{vmatrix} = \hat{i}(0-3) - \hat{j}(-1-6) + \hat{k}(1-0) = -3\hat{i} + 7\hat{j} + \hat{k}$. The equation of the plane is $-3(x+3) + 7(y-7) + 1(z-1) = 0$. Simplifying this, we get $-3x - 9 + 7y - 49 + z - 1 = 0$, which is $-3x + 7y + z - 59 = 0$. The perpendicular distance $p$ from the origin $(0,0,0)$ to the plane $Ax+By+Cz+D=0$ is $p = \frac{|D|}{\sqrt{A^2+B^2+C^2}}$. $p = \frac{|-59|}{\sqrt{(-3)^2 + 7^2 + 1^2}} = \frac{59}{\sqrt{9+49+1}} = \frac{59}{\sqrt{59}} = \sqrt{59}$. Therefore, $\sqrt{p^2+5} = \sqrt{(\sqrt{59})^2 + 5} = \sqrt{59+5} = \sqrt{64} = 8$.

$\vec{n}$ is a unit vector normal to the plane $\pi$ containing the vectors $\hat{i}+3 \hat{k}$ and $2 \hat{i}+\hat{j}-\hat{k}$. If this plane $\pi$ passes through the point $(-3,7,1)$ and $p$ is the perpendicular distance from the origin to this plane $\pi$, then $\sqrt{p^2+5}=$

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