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(A) The line passes through $P(-2, -1, 3)$ and is parallel to $\vec{v} = 3\hat{i} + 2\hat{j} + 2\hat{k}$. Thus,the coordinates of $Q$ can be written a

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$136$

Vedclass · Answer

(A) The line passes through $P(-2, -1, 3)$ and is parallel to $\vec{v} = 3\hat{i} + 2\hat{j} + 2\hat{k}$. Thus,the coordinates of $Q$ can be written as $Q(3\lambda - 2, 2\lambda - 1, 2\lambda + 3)$ for some $\lambda 
eq 0$.Given the distance $QR = 5$,we have $\sqrt{(3\lambda - 2 - 1)^2 + (2\lambda - 1 - 3)^2 + (2\lambda + 3 - 3)^2} = 5$.Squaring both sides: $(3\lambda - 3)^2 + (2\lambda - 4)^2 + (2\lambda)^2 = 25$.$9(\lambda - 1)^2 + 4(\lambda - 2)^2 + 4\lambda^2 = 25$.$9(\lambda^2 - 2\lambda + 1) + 4(\lambda^2 - 4\lambda + 4) + 4\lambda^2 = 25$.$17\lambda^2 - 34\lambda + 25 = 25 \Rightarrow 17\lambda(\lambda - 2) = 0$.Since $Q$ is distinct from $P$,$\lambda 
eq 0$,so $\lambda = 2$.Thus,$Q = (3(2) - 2, 2(2) - 1, 2(2) + 3) = (4, 3, 7)$.Now,$\vec{PQ} = Q - P = (4 - (-2), 3 - (-1), 7 - 3) = (6, 4, 4)$.$\vec{PR} = R - P = (1 - (-2), 3 - (-1), 3 - 3) = (3, 4, 0)$.The area of $	riangle PQR$ is $\frac{1}{2} |\vec{PQ} 	imes \vec{PR}|$.$\vec{PQ} 	imes \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 6 & 4 & 4 \ 3 & 4 & 0 \end{vmatrix} = \hat{i}(0 - 16) - \hat{j}(0 - 12) + \hat{k}(24 - 12) = -16\hat{i} + 12\hat{j} + 12\hat{k}$.$|\vec{PQ} 	imes \vec{PR}| = \sqrt{(-16)^2 + 12^2 + 12^2} = \sqrt{256 + 144 + 144} = \sqrt{544}$.Area $= \frac{1}{2} \sqrt{544} = \sqrt{\frac{544}{4}} = \sqrt{136}$.Therefore,the square of the area is $136$.

Let a line pass through two distinct points $P(-2, -1, 3)$ and $Q$,and be parallel to the vector $3\hat{i} + 2\hat{j} + 2\hat{k}$. If the distance of the point $Q$ from the point $R(1, 3, 3)$ is $5$,then the square of the area of $\triangle PQR$ is equal to:

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