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(B) The plane $P$ passes through points $Q(5,3,0)$,$R(13,3,-2)$,and $S(1,6,2)$.Vectors in the plane are $\vec{QR} = (13-5, 3-3, -2-0) = (8, 0, -2)$ an

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$4$

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(B) The plane $P$ passes through points $Q(5,3,0)$,$R(13,3,-2)$,and $S(1,6,2)$.Vectors in the plane are $\vec{QR} = (13-5, 3-3, -2-0) = (8, 0, -2)$ and $\vec{QS} = (1-5, 6-3, 2-0) = (-4, 3, 2)$.The normal vector $\vec{n} = \vec{QR} 	imes \vec{QS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 8 & 0 & -2 \ -4 & 3 & 2 \end{vmatrix} = \hat{i}(0 - (-6)) - \hat{j}(16 - 8) + \hat{k}(24 - 0) = 6\hat{i} - 8\hat{j} + 24\hat{k}$.Dividing by $2$,we get the normal vector $\vec{n}' = 3\hat{i} - 4\hat{j} + 12\hat{k}$.The equation of the plane is $3(x-5) - 4(y-3) + 12(z-0) = 0$,which simplifies to $3x - 4y + 12z = 3$.The distance of point $A(3,4,\alpha)$ from the plane is $\frac{|3(3) - 4(4) + 12(\alpha) - 3|}{\sqrt{3^2 + (-4)^2 + 12^2}} = \frac{|9 - 16 + 12\alpha - 3|}{13} = \frac{|12\alpha - 10|}{13} = 2$.$|12\alpha - 10| = 26 \implies 12\alpha - 10 = 26$ or $12\alpha - 10 = -26$.$12\alpha = 36 \implies \alpha = 3$ (since $\alpha \in N$).Now,the distance of point $B(2,3,a)$ from the plane is $\frac{|3(2) - 4(3) + 12(a) - 3|}{13} = 3$.$|6 - 12 + 12a - 3| = 39 \implies |12a - 9| = 39$.$12a - 9 = 39 \implies 12a = 48 \implies a = 4$.$12a - 9 = -39 \implies 12a = -30 \implies a = -2.5$.Since we need the positive value,$a = 4$.

Let $P$ be the plane passing through the points $(5,3,0), (13,3,-2)$ and $(1,6,2)$. For $\alpha \in N$,if the distances of the points $A(3,4,\alpha)$ and $B(2,\alpha,a)$ from the plane $P$ are $2$ and $3$ respectively,then the positive value of $a$ is:

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