The distance of the point (-1, 2, -2) from th | vedclass

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(D) Given planes are $P_{1}: 2x + 3y + 2z = 0$ and $P_{2}: x - 2y + z = 0$.The normal vectors are $\vec{n}_{1} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ and $

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$\frac{\sqrt{34}}{2}$

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(D) Given planes are $P_{1}: 2x + 3y + 2z = 0$ and $P_{2}: x - 2y + z = 0$.The normal vectors are $\vec{n}_{1} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ and $\vec{n}_{2} = \hat{i} - 2\hat{j} + \hat{k}$.The direction vector $\vec{v}$ of the line of intersection $L$ is given by $\vec{v} = \vec{n}_{1} 	imes \vec{n}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 2 \ 1 & -2 & 1 \end{vmatrix} = \hat{i}(3 + 4) - \hat{j}(2 - 2) + \hat{k}(-4 - 3) = 7\hat{i} - 7\hat{k}$.Simplifying the direction ratios,we get $(1, 0, -1)$.Since the planes pass through the origin $(0, 0, 0)$,the line $L$ passes through the origin. Thus,the equation of line $L$ is $\frac{x}{1} = \frac{y}{0} = \frac{z}{-1} = \lambda$.Any point $Q$ on the line $L$ is $(\lambda, 0, -\lambda)$.Let $P = (-1, 2, -2)$. The vector $\vec{PQ} = (\lambda + 1, -2, -\lambda + 2)$.Since $\vec{PQ}$ is perpendicular to the line $L$ (direction vector $\vec{v} = (1, 0, -1)$),their dot product is zero:$(\lambda + 1)(1) + (-2)(0) + (-\lambda + 2)(-1) = 0$$\lambda + 1 + \lambda - 2 = 0 \Rightarrow 2\lambda = 1 \Rightarrow \lambda = \frac{1}{2}$.Thus,$Q = (\frac{1}{2}, 0, -\frac{1}{2})$.The distance $PQ = \sqrt{(\frac{1}{2} - (-1))^2 + (0 - 2)^2 + (-\frac{1}{2} - (-2))^2} = \sqrt{(\frac{3}{2})^2 + (-2)^2 + (\frac{3}{2})^2} = \sqrt{\frac{9}{4} + 4 + \frac{9}{4}} = \sqrt{\frac{18}{4} + 4} = \sqrt{\frac{9}{2} + 4} = \sqrt{\frac{17}{2}} = \sqrt{\frac{34}{4}} = \frac{\sqrt{34}}{2}$.

The distance of the point $(-1, 2, -2)$ from the line of intersection of the planes $2x + 3y + 2z = 0$ and $x - 2y + z = 0$ is:

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