The plane 2x - y + z = 4 intersects the line | vedclass

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(A) Given points $A(a, -2, 4)$ and $B(2, b, -3)$.Point $C$ divides $AB$ in the ratio $2:1$. Using the section formula,the coordinates of $C$ are:$C =

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$\frac{17}{3}$

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(A) Given points $A(a, -2, 4)$ and $B(2, b, -3)$.Point $C$ divides $AB$ in the ratio $2:1$. Using the section formula,the coordinates of $C$ are:$C = \left( \frac{2(2) + 1(a)}{2+1}, \frac{2(b) + 1(-2)}{2+1}, \frac{2(-3) + 1(4)}{2+1} \right) = \left( \frac{a+4}{3}, \frac{2b-2}{3}, \frac{-2}{3} \right)$.Since $C$ lies on the plane $2x - y + z = 4$:$2\left( \frac{a+4}{3} \right) - \left( \frac{2b-2}{3} \right) + \left( \frac{-2}{3} \right) = 4$$2a + 8 - 2b + 2 - 2 = 12 \Rightarrow 2a - 2b = 4 \Rightarrow a - b = 2 \Rightarrow a = b + 2$.Given the distance $OC = \sqrt{5}$,so $OC^2 = 5$:$\left( \frac{a+4}{3} \right)^2 + \left( \frac{2b-2}{3} \right)^2 + \left( \frac{-2}{3} \right)^2 = 5$$(b+2+4)^2 + (2b-2)^2 + 4 = 45$$(b+6)^2 + (2b-2)^2 = 41$$b^2 + 12b + 36 + 4b^2 - 8b + 4 = 41$$5b^2 + 4b - 1 = 0 \Rightarrow (5b - 1)(b + 1) = 0$.So,$b = -1$ or $b = 1/5$. Since $ab  0$.Thus,$a = 1, b = -1$.$C = \left( \frac{1+4}{3}, \frac{-2-2}{3}, \frac{-2}{3} \right) = \left( \frac{5}{3}, -\frac{4}{3}, -\frac{2}{3} \right)$.$P = (a-b, b, 2b-a) = (1 - (-1), -1, 2(-1) - 1) = (2, -1, -3)$.$CP^2 = \left( 2 - \frac{5}{3} \right)^2 + \left( -1 - (-\frac{4}{3}) \right)^2 + \left( -3 - (-\frac{2}{3}) \right)^2$$CP^2 = \left( \frac{1}{3} \right)^2 + \left( \frac{1}{3} \right)^2 + \left( -\frac{7}{3} \right)^2 = \frac{1}{9} + \frac{1}{9} + \frac{49}{9} = \frac{51}{9} = \frac{17}{3}$.

The plane $2x - y + z = 4$ intersects the line segment joining the points $A(a, -2, 4)$ and $B(2, b, -3)$ at the point $C$ in the ratio $2:1$. The distance of the point $C$ from the origin is $\sqrt{5}$. If $ab < 0$ and $P$ is the point $(a - b, b, 2b - a)$,then $CP^2$ is equal to:

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