The vertices of a tetrahedron are A(3, 4, 2), B(1, 2, 1), C(4, 1, 3), D( $-$ 1, $-$ 1,3). Find the height of A above the base BCD is $\frac{a}{\sqrt{b}}$ then sum of the digits of b is k then $\frac{k}{6} =$

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Detailed Solution

The equation of the plane containing $BCD = |\begin{matrix} x - 1 & y - 2 & z - 1 \\ 3 & - 1 & 2 \\ - 2 & - 3 & 2 \end{matrix}| = 0$

$\Rightarrow 4 x - 10 y - 11 z + 27 = 0 ∴$ distance from $A (3, 4, 2) = \frac{|4 (3) - 10 (4) - 11 (2) + 27|}{\sqrt{16 + 100 + 121}} \frac{23}{\sqrt{237}}$

b=237

k=sum of the digits of b=2+3+7=12

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