Line Parallel to Given Line Intersecting Planes at Points P Q R

To determine which of the given statements are true, we break down the problem into step-by-step geometric calculations.

Step 1: Find the coordinates of point $Q$

The given point is $P(1, 3, 2)$. The straight line through $P$ is parallel to the line: $\frac{x-2}{1} = \frac{y-4}{2} = \frac{z-6}{1}$

Thus, its direction vector is $\vec{d} = \langle 1, 2, 1 \rangle$.

The equation of the line passing through $P(1, 3, 2)$ is: $\frac{x-1}{1} = \frac{y-3}{2} = \frac{z-2}{1} = \lambda$

Any general point on this line can be expressed as: $Q_\lambda = (1+\lambda, \, 3+2\lambda, \, 2+\lambda)$

Since this line intersects the plane $L_1 : x - y + 3z = 6$ at $Q$, we substitute $Q_\lambda$ into the equation of $L_1$: $(1+\lambda) - (3+2\lambda) + 3(2+\lambda) = 6$ $1 + \lambda - 3 - 2\lambda + 6 + 3\lambda = 6$ $2\lambda + 4 = 6 \implies 2\lambda = 2 \implies \lambda = 1$

Substituting $\lambda = 1$ back into the coordinates of $Q_\lambda$: $Q = (1+1, \, 3+2(1), \, 2+1) = (2, 5, 3)$

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Step 2: Find the length of line segment $PQ$

Using the distance formula between $P(1, 3, 2)$ and $Q(2, 5, 3)$: $PQ = \sqrt{(2-1)^2 + (5-3)^2 + (3-2)^2} = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}$

Thus, Option (A) is TRUE.

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Step 3: Find the coordinates of point $R$

A line passing through $Q(2, 5, 3)$ is perpendicular to plane $L_1 : x - y + 3z = 6$. The normal vector to plane $L_1$ is $\vec{n} = \langle 1, -1, 3 \rangle$.

The equation of this perpendicular line is: $\frac{x-2}{1} = \frac{y-5}{-1} = \frac{z-3}{3} = \mu$

Any general point on this line can be written as: $R_\mu = (2+\mu, \, 5-\mu, \, 3+3\mu)$

Since this line intersects the plane $L_2 : 2x - y + z = -4$ at $R$, we substitute $R_\mu$ into $L_2$: $2(2+\mu) - (5-\mu) + (3+3\mu) = -4$ $4 + 2\mu - 5 + \mu + 3 + 3\mu = -4$ $6\mu + 2 = -4 \implies 6\mu = -6 \implies \mu = -1$

Substituting $\mu = -1$ gives the coordinates of $R$: $R = (2 + (-1), \, 5 - (-1), \, 3 + 3(-1)) = (1, 6, 0)$

Since the coordinates of $R$ are $(1, 6, 0)$, Option (B) is FALSE.

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Step 4: Find the centroid of $\triangle PQR$

The centroid $G$ of $\triangle PQR$ with vertices $P(1,3,2)$, $Q(2,5,3)$, and $R(1,6,0)$ is given by: $G = \left( \frac{1+2+1}{3}, \, \frac{3+5+6}{3}, \, \frac{2+3+0}{3} \right) = \left( \frac{4}{3}, \, \frac{14}{3}, \, \frac{5}{3} \right)$

Thus, Option (C) is TRUE.

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Step 5: Find the perimeter of $\triangle PQR$

We calculate the side lengths of $\triangle PQR$:

• $PQ = \sqrt{6}$
• $QR = \sqrt{(1-2)^2 + (6-5)^2 + (0-3)^2} = \sqrt{(-1)^2 + 1^2 + (-3)^2} = \sqrt{11}$
• $PR = \sqrt{(1-1)^2 + (6-3)^2 + (0-2)^2} = \sqrt{0^2 + 3^2 + (-2)^2} = \sqrt{13}$

The perimeter of $\triangle PQR$ is: $\text{Perimeter} = PQ + QR + PR = \sqrt{6} + \sqrt{11} + \sqrt{13}$

Thus, Option (D) is FALSE.

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Conclusion

The TRUE statements are (A) and (C).