If $$\lambda_1 < \lambda_2$$ are two values of $$\lambda$$ such that the angle between the planes $$P_1: \vec{r} \cdot (3\hat{i} - 5\hat{j} + \hat{k}) = 7$$ and $$P_2: \vec{r} \cdot (\lambda\hat{i} + \hat{j} - 3\hat{k}) = 9$$ is $$\sin^{-1}\frac{2\sqrt{6}}{5}$$, then the square of the length of perpendicular from the point $$(38\lambda_1, 10\lambda_2, 2)$$ to the plane $$P_1$$ is

To find the required square of the length of the perpendicular, we first determine the values of $$\lambda$$ using the given angle between the two planes.

The normal vectors to the planes $$P_1$$ and $$P_2$$ are:

$$\vec{n}_1 = 3\hat{i} - 5\hat{j} + \hat{k}$$

$$\vec{n}_2 = \lambda\hat{i} + \hat{j} - 3\hat{k}$$

The angle $$\theta$$ between the two planes is given by:

$$\theta = \sin^{-1}\left(\frac{2\sqrt{6}}{5}\right) \implies \sin\theta = \frac{2\sqrt{6}}{5}$$

Using the identity $$\cos^2\theta = 1 - \sin^2\theta$$:

$$\cos^2\theta = 1 - \left(\frac{2\sqrt{6}}{5}\right)^2 = 1 - \frac{24}{25} = \frac{1}{25}$$

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Step 1: Apply the angle formula between two planes

The cosine of the angle between two normal vectors is given by:

$$\cos\theta = \frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$$

Computing the dot product and the magnitudes:

$$\vec{n}_1 \cdot \vec{n}_2 = 3(\lambda) + (-5)(1) + (1)(-3) = 3\lambda - 8$$

$$|\vec{n}_1| = \sqrt{3^2 + (-5)^2 + 1^2} = \sqrt{9 + 25 + 1} = \sqrt{35}$$

$$|\vec{n}_2| = \sqrt{\lambda^2 + 1^2 + (-3)^2} = \sqrt{\lambda^2 + 10}$$

Squaring both sides of the relation:

$$\cos^2\theta = \frac{(3\lambda - 8)^2}{35(\lambda^2 + 10)} = \frac{1}{25}$$

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Step 2: Solve for $$\lambda$$

Cross-multiplying the terms:

$$25(3\lambda - 8)^2 = 35(\lambda^2 + 10)$$

$$5(9\lambda^2 - 48\lambda + 64) = 7(\lambda^2 + 10)$$

$$45\lambda^2 - 240\lambda + 320 = 7\lambda^2 + 70$$

$$38\lambda^2 - 240\lambda + 250 = 0$$

Dividing the entire equation by 2:

$$19\lambda^2 - 120\lambda + 125 = 0$$

Using the quadratic formula to find the roots:

$$\lambda = \frac{120 \pm \sqrt{(-120)^2 - 4(19)(125)}}{2(19)}$$

$$\lambda = \frac{120 \pm \sqrt{14400 - 9500}}{38} = \frac{120 \pm \sqrt{4900}}{38} = \frac{120 \pm 70}{38}$$

Thus, the two values of $$\lambda$$ are:

$$\lambda_1 = \frac{120 - 70}{38} = \frac{50}{38} = \frac{25}{19}$$

$$\lambda_2 = \frac{120 + 70}{38} = \frac{190}{38} = 5$$

Given that $$\lambda_1 < \lambda_2$$, we assign:

$$\lambda_1 = \frac{25}{19} \quad \text{and} \quad \lambda_2 = 5$$

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Step 3: Find the coordinates of the point

Substituting the values of $$\lambda_1$$ and $$\lambda_2$$ into the given point $$(38\lambda_1, 10\lambda_2, 2)$$:

$$\text{Point} = \left(38 \times \frac{25}{19}, 10 \times 5, 2\right) = (50, 50, 2)$$

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Step 4: Calculate the square of the perpendicular length to $$P_1$$

The Cartesian equation of the plane $$P_1$$ is:

$$3x - 5y + z - 7 = 0$$

The perpendicular distance $$d$$ from a point $$(x_1, y_1, z_1)$$ to the plane $$Ax + By + Cz + D = 0$$ is given by:

$$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

Substituting our point $$(50, 50, 2)$$ into the formula:

$$d = \frac{|3(50) - 5(50) + 2 - 7|}{\sqrt{3^2 + (-5)^2 + 1^2}}$$

$$d = \frac{|150 - 250 - 5|}{\sqrt{35}} = \frac{|-105|}{\sqrt{35}} = \frac{105}{\sqrt{35}}$$

Squaring the distance gives:

$$d^2 = \frac{105^2}{35} = \frac{11025}{35} = 315$$

Therefore, the square of the length of the perpendicular is equal to 315.