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Question

Let the line (x-3 / 7)=(y-2 / -1)=(z-3 / -4) intersect the plane containing the lines (x-4 / 1)=(y+1 / -2)=(z / 1) and 4 a x-y+5 z-7 a=0=2 x-5 y-z-3, a in mathbbR at the point P(alpha, beta, gamma). Then the value of alpha+beta+gamma equals _____________.

DivineJEE · Accepted Answer

Equation of plane containing the line 4ax - y + 5z - 7a = 0 = 2x - 5y - z - 3 can be written as 4ax - y + 5z - 7a + lambda (2x - 5y - z - 3) = 0 (4a + 2lambda )x - (1 + 5lambda )y \ + (5 - lambda )z - (7z + 3lambda ) = 0 Which is coplanar with the line x - 4 / 1 = y + 1 / - 2 = z / 1 4(4a + 2lambda ) + (1 + 5lambda ) - (7a + 3lambda ) = 0 9a + 10lambda + 1 = 0 ..... (1) (4a + 2lambda )1 + (1 + 5lambda )2 + 5 - lambda = 0 4a + 11lambda + 7 = 0 ...... (2) a = 1, lambda = - 1 Equation of plane is x + 2y + 3z - 2 = 0 Intersection with the line x - 3 / 7 = y - 2 / - 1 = z - 3 / - 4 (7t + 3) + 2( - t + 2) + 3( - 4t + 3) - 2 = 0 - 7t + 14 = 0 t = 2 So, the required point is (17,0, - 5) alpha + beta + gamma = 12

Explanation

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