Let the line x-37=y-2-1=z-3-4 intersect the plane containing the lines x-41=y+1-2=z1 and 4 a x-y+5 z-7 a=0=2 x-5 y-z-3, a R at the point P(, , ). Then the value of ++ equals _____________.

Equation of plane containing the line 4ax - y + 5z - 7a = 0 = 2x - 5y - z - 3 can be written as 4ax - y + 5z - 7a + (2x - 5y - z - 3) = 0 (4a + 2 )x - (1 + 5 )y \\\\ + (5 - )z - (7z + 3 ) = 0 Which is coplanar with the line x - 4 1 = y + 1 - 2 = z 1 4(4a + 2 ) + (1 + 5 ) - (7a + 3 ) = 0 9a + 10 + 1 = 0 .

Let the line frac x37frac y21frac z34 intersect the plane co: null (Mathematics)

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 + 2\lambda )x - (1 + 5\lambda )y \\\\ + (5 - \lambda )z - (7z + 3\lambda ) = 0

Which is coplanar with the line

{{x - 4} \over 1} = {{y + 1} \over { - 2}} = {z \over 1}

4(4a + 2\lambda ) + (1 + 5\lambda ) - (7a + 3\lambda ) = 0

9a + 10\lambda + 1 = 0 \quad ..... (1)

(4a + 2\lambda )1 + (1 + 5\lambda )2 + 5 - \lambda = 0

4a + 11\lambda + 7 = 0 \quad ...... (2)

a = 1,\,\lambda = - 1

Equation of plane is

x + 2y + 3z - 2 = 0

Intersection with the line

{{x - 3} \over 7} = {{y - 2} \over { - 1}} = {{z - 3} \over { - 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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