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(B) The equation of the plane passing through points $A(1, 2, 3)$,$B(2, 3, -4)$,and $C(3, -4, 5)$ is given by the determinant equation: $\left|\begin{

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$\frac{12}{\sqrt{30}}$

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(B) The equation of the plane passing through points $A(1, 2, 3)$,$B(2, 3, -4)$,and $C(3, -4, 5)$ is given by the determinant equation: $\left|\begin{array}{ccc} x-1 & y-2 & z-3 \ 2-1 & 3-2 & -4-3 \ 3-1 & -4-2 & 5-3 \end{array}ight| = 0$ $\Rightarrow \left|\begin{array}{ccc} x-1 & y-2 & z-3 \ 1 & 1 & -7 \ 2 & -6 & 2 \end{array}ight| = 0$ Expanding along the first row: $(x-1)(2 - 42) - (y-2)(2 - (-14)) + (z-3)(-6 - 2) = 0$ $(x-1)(-40) - (y-2)(16) + (z-3)(-8) = 0$ $-40x + 40 - 16y + 32 - 8z + 24 = 0$ $-40x - 16y - 8z + 96 = 0$ Dividing by $-8$: $5x + 2y + z - 12 = 0$ The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$. $d = \frac{|-12|}{\sqrt{5^2 + 2^2 + 1^2}} = \frac{12}{\sqrt{25 + 4 + 1}} = \frac{12}{\sqrt{30}}$.

The perpendicular distance from the origin to the plane containing the points having position vectors $\hat{i}+2\hat{j}+3\hat{k}$,$2\hat{i}+3\hat{j}-4\hat{k}$,and $3\hat{i}-4\hat{j}+5\hat{k}$ is

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