Let Q and R be two points on the line frac{x+ | vedclass

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(A) Let any point on the line be $A(\lambda) = (2\lambda - 1, 3\lambda - 2, 2\lambda + 1)$.Given the distance from $P(4, 2, 7)$ is $\sqrt{26}$,we have

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$153$

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(A) Let any point on the line be $A(\lambda) = (2\lambda - 1, 3\lambda - 2, 2\lambda + 1)$.Given the distance from $P(4, 2, 7)$ is $\sqrt{26}$,we have:$(2\lambda - 1 - 4)^2 + (3\lambda - 2 - 2)^2 + (2\lambda + 1 - 7)^2 = (\sqrt{26})^2$$(2\lambda - 5)^2 + (3\lambda - 4)^2 + (2\lambda - 6)^2 = 26$$(4\lambda^2 - 20\lambda + 25) + (9\lambda^2 - 24\lambda + 16) + (4\lambda^2 - 24\lambda + 36) = 26$$17\lambda^2 - 68\lambda + 77 = 26$$17\lambda^2 - 68\lambda + 51 = 0$$\lambda^2 - 4\lambda + 3 = 0 \implies (\lambda - 1)(\lambda - 3) = 0$So,$\lambda = 1$ and $\lambda = 3$.For $\lambda = 1$,$Q = (1, 1, 3)$. For $\lambda = 3$,$R = (5, 7, 7)$.$\overrightarrow{PQ} = Q - P = (1-4, 1-2, 3-7) = (-3, -1, -4)$.$\overrightarrow{PR} = R - P = (5-4, 7-2, 7-7) = (1, 5, 0)$.$\overrightarrow{PQ} 	imes \overrightarrow{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & -1 & -4 \ 1 & 5 & 0 \end{vmatrix} = \hat{i}(0 - (-20)) - \hat{j}(0 - (-4)) + \hat{k}(-15 - (-1)) = 20\hat{i} - 4\hat{j} - 14\hat{k}$.Area of $	riangle PQR = \frac{1}{2} |\overrightarrow{PQ} 	imes \overrightarrow{PR}| = \frac{1}{2} \sqrt{20^2 + (-4)^2 + (-14)^2} = \frac{1}{2} \sqrt{400 + 16 + 196} = \frac{1}{2} \sqrt{612} = \sqrt{\frac{612}{4}} = \sqrt{153}$.The square of the area is $(\sqrt{153})^2 = 153$.

Let $Q$ and $R$ be two points on the line $\frac{x+1}{2} = \frac{y+2}{3} = \frac{z-1}{2}$ at a distance of $\sqrt{26}$ from the point $P(4, 2, 7)$. Then the square of the area of the triangle $PQR$ is $....$

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