Let P_1 and P_2 be two planes given by P_1: 1 | vedclass

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(A) The line of intersection of the two planes $P_1$ and $P_2$ is found by solving the system of equations:$10x + 15y + 12z = 60$$-2x + 5y + 4z = 20$M

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$A, B, C$

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(A) The line of intersection of the two planes $P_1$ and $P_2$ is found by solving the system of equations:$10x + 15y + 12z = 60$$-2x + 5y + 4z = 20$Multiplying the second equation by $5$,we get $-10x + 25y + 20z = 100$. Adding this to the first equation gives $40y + 32z = 160$,or $5y + 4z = 20$. Setting $z = 5k$,we get $5y = 20 - 20k$,so $y = 4 - 4k$. Substituting into the second plane equation: $-2x + 5(4 - 4k) + 4(5k) = 20 \implies -2x + 20 - 20k + 20k = 20 \implies x = 0$.The line of intersection is $\frac{x}{0} = \frac{y-4}{-4} = \frac{z}{5}$.An edge of a tetrahedron whose two faces lie on $P_1$ and $P_2$ must either be skew to the line of intersection or intersect it at a point not on the planes (or lie within one of the planes). Checking the options,lines $A$,$B$,and $C$ satisfy the conditions to be edges of such a tetrahedron.

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