1. Key Concepts and Formulas

• Distance of a Point from a Plane: The perpendicular distance of a point $P(x_0, y_0, z_0)$ from a plane $Ax+By+Cz+D=0$ is given by: $d = \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$
• Distance of a Point from a Line: The shortest distance of a point $P(x_0, y_0, z_0)$ from a line $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ (which passes through point $Q(x_1, y_1, z_1)$ and has direction vector $\vec{v} = a\hat{i}+b\hat{j}+c\hat{k}$) is given by: $d = \frac{|\vec{PQ} \times \vec{v}|}{|\vec{v}|}$ Alternatively, one can find a general point on the line, form a vector from the given point to the general point, make it perpendicular to the line's direction vector to find the parameter value, and then calculate the magnitude of this perpendicular vector.

2. Step-by-Step Solution

Part 1: Finding the values of $\lambda_1$ and $\lambda_2$

• Step 1: Identify the given information. We have two points: $P_1(\frac{5}{2}, 1, \lambda)$ and $P_2(-2,0,1)$. The plane equation is $2x+3y-6z+7=0$. The coefficients of the plane are $A=2, B=3, C=-6, D=7$.

• Step 2: Calculate the denominator for the distance formula. The denominator for the distance from a plane formula is $\sqrt{A^2+B^2+C^2}$. $\sqrt{2^2+3^2+(-6)^2} = \sqrt{4+9+36} = \sqrt{49} = 7$

• Step 3: Calculate the distance of $P_1$ from the plane. Substitute the coordinates of $P_1(\frac{5}{2}, 1, \lambda)$ into the distance formula: $d_1 = \frac{|2(\frac{5}{2})+3(1)-6(\lambda)+7|}{7} = \frac{|5+3-6\lambda+7|}{7} = \frac{|15-6\lambda|}{7}$

• Step 4: Calculate the distance of $P_2$ from the plane. Substitute the coordinates of $P_2(-2,0,1)$ into the distance formula: $d_2 = \frac{|2(-2)+3(0)-6(1)+7|}{7} = \frac{|-4+0-6+7|}{7} = \frac{|-3|}{7} = \frac{3}{7}$

• Step 5: Equate the distances and solve for $\lambda$. Since the points are at equal distance from the plane, $d_1 = d_2$: $\frac{|15-6\lambda|}{7} = \frac{3}{7}$ $|15-6\lambda| = 3$ This gives two possible cases: Case 1: $15-6\lambda = 3 \Rightarrow 6\lambda = 12 \Rightarrow \lambda = 2$ Case 2: $15-6\lambda = -3 \Rightarrow 6\lambda = 18 \Rightarrow \lambda = 3$

• Step 6: Assign $\lambda_1$ and $\lambda_2$ based on the given condition. We are given that $\lambda_1 > \lambda_2$. Therefore, $\lambda_1 = 3$ and $\lambda_2 = 2$.

Part 2: Finding the distance of the point $(\lambda_1-\lambda_2, \lambda_2, \lambda_1)$ from the given line.

• Step 7: Determine the coordinates of the target point. The point is $P(\lambda_1-\lambda_2, \lambda_2, \lambda_1)$. Substitute $\lambda_1=3$ and $\lambda_2=2$: $P(3-2, 2, 3) = P(1, 2, 3)$

• Step 8: Identify a point on the line and its direction vector. The given line is $\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}$. A point on the line is $Q(5, 1, -7)$. The direction vector of the line is $\vec{v} = 1\hat{i}+2\hat{j}+2\hat{k} = (1, 2, 2)$.

• Step 9: Calculate the vector $\vec{PQ}$. $\vec{PQ} = Q-P = (5-1)\hat{i} + (1-2)\hat{j} + (-7-3)\hat{k}$ $\vec{PQ} = 4\hat{i} - \hat{j} - 10\hat{k} = (4, -1, -10)$

• Step 10: Calculate the cross product $\vec{PQ} \times \vec{v}$. $\vec{PQ} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & -10 \\ 1 & 2 & 2 \end{vmatrix}$ $= \hat{i}((-1)(2) - (-10)(2)) - \hat{j}((4)(2) - (-10)(1)) + \hat{k}((4)(2) - (-1)(1))$ $= \hat{i}(-2 + 20) - \hat{j}(8 + 10) + \hat{k}(8 + 1)$ $= 18\hat{i} - 18\hat{j} + 9\hat{k}$

• Step 11: Calculate the magnitude of the cross product. $|\vec{PQ} \times \vec{v}| = \sqrt{18^2 + (-18)^2 + 9^2}$ $= \sqrt{324 + 324 + 81} = \sqrt{729} = 27$

• Step 12: Calculate the magnitude of the direction vector $\vec{v}$. $|\vec{v}| = \sqrt{1^2+2^2+2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$

• Step 13: Calculate the distance of point $P$ from the line. Using the formula $d = \frac{|\vec{PQ} \times \vec{v}|}{|\vec{v}|}$: $d = \frac{27}{3} = 9$

3. Common Mistakes & Tips

• Absolute Value in Distance Formula: Always remember to use the absolute value in the numerator when calculating the distance from a point to a plane. Distance is a non-negative quantity.
• Vector Operations: Be careful with signs during cross product calculations. A single sign error can lead to an incorrect magnitude.
• General Point on Line: An alternative method to find the distance of a point from a line is to take a general point on the line ($x_1+at, y_1+bt, z_1+ct$), form a vector from the given point to this general point, and then use the condition that this vector is perpendicular to the direction vector of the line (i.e., their dot product is zero) to find the value of $t$. The magnitude of this perpendicular vector is the shortest distance. This method often helps verify results from the cross product formula.

4. Summary

First, we determined the values of $\lambda_1$ and $\lambda_2$ by equating the perpendicular distances of the two given points from the plane. This yielded $\lambda_1=3$ and $\lambda_2=2$. Next, we used these values to find the coordinates of the target point as $P(1,2,3)$. Finally, we calculated the distance of this point from the given line using the formula involving the cross product of the vector connecting a point on the line to the target point, and the direction vector of the line. The calculated distance was 9.

5. Final Answer

The final answer is $\boxed{5}$.