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(A) The equation of the plane $P$ passing through the intersection of $P_1: 2x + 3y - z - 2 = 0$ and $P_2: x + 2y + 3z - 6 = 0$ is given by $P_1 + \la

Vedclass · Accepted Answer

$\frac{250}{83}$

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(A) The equation of the plane $P$ passing through the intersection of $P_1: 2x + 3y - z - 2 = 0$ and $P_2: x + 2y + 3z - 6 = 0$ is given by $P_1 + \lambda P_2 = 0$.$(2x + 3y - z - 2) + \lambda(x + 2y + 3z - 6) = 0$$(2 + \lambda)x + (3 + 2\lambda)y + (3\lambda - 1)z - (2 + 6\lambda) = 0$.Since $P$ is perpendicular to the plane $2x + y - z + 1 = 0$,the dot product of their normals is zero:$2(2 + \lambda) + 1(3 + 2\lambda) - 1(3\lambda - 1) = 0$$4 + 2\lambda + 3 + 2\lambda - 3\lambda + 1 = 0$$\lambda + 8 = 0 \implies \lambda = -8$.Substituting $\lambda = -8$ into the equation of $P$:$(2 - 8)x + (3 - 16)y + (-24 - 1)z - (2 - 48) = 0$$-6x - 13y - 25z + 46 = 0 \implies 6x + 13y + 25z - 46 = 0$.The distance $d$ from the point $(-7, 1, 1)$ to the plane $6x + 13y + 25z - 46 = 0$ is:$d = \frac{|6(-7) + 13(1) + 25(1) - 46|}{\sqrt{6^2 + 13^2 + 25^2}} = \frac{|-42 + 13 + 25 - 46|}{\sqrt{36 + 169 + 625}} = \frac{|-50|}{\sqrt{830}} = \frac{50}{\sqrt{830}}$.Therefore,$d^2 = \frac{50^2}{830} = \frac{2500}{830} = \frac{250}{83}$.

Let the plane $P$ pass through the intersection of the planes $2x + 3y - z = 2$ and $x + 2y + 3z = 6$,and be perpendicular to the plane $2x + y - z + 1 = 0$. If $d$ is the distance of $P$ from the point $(-7, 1, 1)$,then $d^2$ is equal to:

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