Let the foot of perpendicular of the point $P(3,-2,-9)$ on the plane passing through the points $(-1,-2,-3),(9,3,4),(9,-2,1)$ be $Q(\alpha, \beta, \gamma)$. Then the distance of $Q$ from the origin is :

<p>The equation of the plane passing through points $A(-1, -2, -3)$, $B(9, 3, 4)$, and $C(9, -2, 1)$ can be written using the determinant :</p> <p>$$ \left|\begin{array}{ccc} x+1 & y+2 & z+3 \\ 10 & 5 & 7 \\ 10 & 0 & 4 \end{array}\right|=0 $$</p> <p>Expanding the determinant, we get :</p> <p>$2x + 3y - 5z - 7 = 0$</p> <p>Next, we find the foot of the perpendicular from point $P(3, -2, -9)$ to the plane. Using the coordinates of $P$ and the equation of the plane, we can find the ratio of the perpendicular distance to the sum of the squares of the coefficients of $x$, $y$, and $z$ :</p> <p>$\frac{2(3) + 3(-2) - 5(-9) - 7}{2^2 + 3^2 + (-5)^2} = \frac{-38}{38}$</p> <p>We can now find the coordinates of the foot of the perpendicular, $Q(\alpha, \beta, \gamma)$ :</p> <p>$$ \frac{\alpha - 3}{2} = \frac{\beta + 2}{3} = \frac{\gamma + 9}{-5} = -\frac{38}{38} $$</p> <p>Solving for $\alpha$, $\beta$, and $\gamma$ :</p> <p>$\alpha = 3 - 2\left(-\frac{38}{38}\right) = 1$</p> <p>$\beta = -2 + 3\left(-\frac{38}{38}\right) = -5$</p> <p>$\gamma = -9 - 5\left(-\frac{38}{38}\right) = -4$</p> <p>So, the coordinates of the foot of the perpendicular are $Q(1, -5, -4)$. Now, we can find the distance of point $Q$ from the origin :</p> <p>$$ OQ = \sqrt{\alpha^2 + \beta^2 + \gamma^2} = \sqrt{1^2 + (-5)^2 + (-4)^2} = \sqrt{42} $$</p>