A plane pi passes through the points (5,1,2), | vedclass

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(C) Let the points be $A(5,1,2)$, $B(3,-4,6)$, and $C(7,0,-1)$.The equation of the plane passing through these points is given by $\begin{vmatrix} x-5

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$p$

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(C) Let the points be $A(5,1,2)$, $B(3,-4,6)$, and $C(7,0,-1)$.The equation of the plane passing through these points is given by $\begin{vmatrix} x-5 & y-1 & z-2 \ 3-5 & -4-1 & 6-2 \ 7-5 & 0-1 & -1-2 \end{vmatrix} = 0$.$\begin{vmatrix} x-5 & y-1 & z-2 \ -2 & -5 & 4 \ 2 & -1 & -3 \end{vmatrix} = 0$.$(x-5)(15+4) - (y-1)(6-8) + (z-2)(2+10) = 0$.$19(x-5) + 2(y-1) + 12(z-2) = 0$.$19x - 95 + 2y - 2 + 12z - 24 = 0$.$19x + 2y + 12z - 121 = 0$.The normal vector is $\vec{n} = 19\hat{i} + 2\hat{j} + 12\hat{k}$.The magnitude is $|\vec{n}| = \sqrt{19^2 + 2^2 + 12^2} = \sqrt{361 + 4 + 144} = \sqrt{509}$.The direction cosines are $l = \frac{19}{\sqrt{509}}$, $m = \frac{2}{\sqrt{509}}$, $n = \frac{12}{\sqrt{509}}$.The perpendicular distance $p$ from the origin $(0,0,0)$ to the plane $19x + 2y + 12z - 121 = 0$ is $p = \frac{|-121|}{\sqrt{509}} = \frac{121}{\sqrt{509}}$.Now, $|3l + 2m + 5n| = |3(\frac{19}{\sqrt{509}}) + 2(\frac{2}{\sqrt{509}}) + 5(\frac{12}{\sqrt{509}})| = |\frac{57 + 4 + 60}{\sqrt{509}}| = \frac{121}{\sqrt{509}} = p$.

$A$ plane $\pi$ passes through the points $(5,1,2)$, $(3,-4,6)$, and $(7,0,-1)$. If $p$ is the perpendicular distance from the origin to the plane $\pi$ and $l, m, n$ are the direction cosines of a normal to the plane $\pi$, then $|3l+2m+5n|=$

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