Let lambda_1, lambda_2 be the values of lambd | vedclass

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(B) The distance of a point $(x_0, y_0, z_0)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 +

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$9$

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(B) The distance of a point $(x_0, y_0, z_0)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.For the point $\left(\frac{5}{2}, 1, \lambdaight)$ and the plane $2x + 3y - 6z + 7 = 0$:$d_1 = \frac{|2(\frac{5}{2}) + 3(1) - 6(\lambda) + 7|}{\sqrt{2^2 + 3^2 + (-6)^2}} = \frac{|5 + 3 - 6\lambda + 7|}{\sqrt{4 + 9 + 36}} = \frac{|15 - 6\lambda|}{7}$.For the point $(-2, 0, 1)$ and the plane $2x + 3y - 6z + 7 = 0$:$d_2 = \frac{|2(-2) + 3(0) - 6(1) + 7|}{\sqrt{4 + 9 + 36}} = \frac{|-4 - 6 + 7|}{7} = \frac{|-3|}{7} = \frac{3}{7}$.Since $d_1 = d_2$,we have $\frac{|15 - 6\lambda|}{7} = \frac{3}{7}$,which implies $|15 - 6\lambda| = 3$.This gives two cases:$15 - 6\lambda = 3 \Rightarrow 6\lambda = 12 \Rightarrow \lambda = 2$.$15 - 6\lambda = -3 \Rightarrow 6\lambda = 18 \Rightarrow \lambda = 3$.Given $\lambda_1 > \lambda_2$,we have $\lambda_1 = 3$ and $\lambda_2 = 2$.The point is $(\lambda_1 - \lambda_2, \lambda_2, \lambda_1) = (3 - 2, 2, 3) = (1, 2, 3)$.The distance of point $P(1, 2, 3)$ from the line $\frac{x - 5}{1} = \frac{y - 1}{2} = \frac{z + 7}{2}$ is given by $d = \frac{|\vec{AP} 	imes \vec{v}|}{|\vec{v}|}$,where $A(5, 1, -7)$ is a point on the line and $\vec{v} = (1, 2, 2)$ is the direction vector.$\vec{AP} = (1 - 5, 2 - 1, 3 - (-7)) = (-4, 1, 10)$.$\vec{AP} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -4 & 1 & 10 \ 1 & 2 & 2 \end{vmatrix} = \hat{i}(2 - 20) - \hat{j}(-8 - 10) + \hat{k}(-8 - 1) = -18\hat{i} + 18\hat{j} - 9\hat{k}$.$|\vec{AP} 	imes \vec{v}| = \sqrt{(-18)^2 + 18^2 + (-9)^2} = \sqrt{324 + 324 + 81} = \sqrt{729} = 27$.$|\vec{v}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$.$d = \frac{27}{3} = 9$.

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