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Find a plane containing the point \( (-1,2,2) \) and the line of intersection of the planes \( 3 x+2 y-3 z=-13 \) and \( 3 x-3 y-2 z=7 \). The equation of the plane is:
Given point \( A(7,5,-2) \) and plane \( p \) with the equation \( 2 x+5 y+9 z=83 \). a. Point \( A \) on the plane \( p \). b. Find the normal vector \( \vec{n} \) to the plane \( p \). c. Find point B on plane \( p \) and \( z \)-axis. The coordinates of point \( B \) are d. \( |\overrightarrow{A B} \cdot \vec{n}|= \) e. Use the formula \( d=\frac{|\overrightarrow{A B} \cdot \vec{n}|}{|\vec{n}|} \) to find the distance \( d \) from the point A to the plane \( p \). \( d= \)