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The plane passing through the intersection fo the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to y-axis also passes through the point:
Option 1)
$(-3,0,-1)$
Option 2)
$(-3,1,1)$
Option 3)
$(3,3,-1)$
Option 4)
$(3,2,1)$

Answers (1)

best_answer

Equation of any plane passing through the line of intersection of two planes (Cartesian form ) -

The equation of any plane passing through the line of intersection of two planes

$ax+by+cz+d= 0$ and

$a_{1}x+b_{1}y+c_{1}z+d_{1}= 0$ is given by

$\left ( ax+by+cz+d \right )+\lambda \left ( a_{1}x+b_{1}y+c_{1}z+d _{1}\right )= 0$

-

As we have learnt from the concept Equation of plane

$\left ( x+y+z-1 \right )+\lambda \left ( 2x+3y-z+4 \right )=0$

$\Rightarrow \left ( 1+2\lambda \right )x+\left ( 1+3\lambda \right )y+\left ( 1-\lambda \right )z-1+4\lambda =0$

Direction ratio of Normal of the plane are ....

$1+2\lambda ,\: \: \: 1+3\lambda \lambda \: \: \: \: \: \: \: \: and\: \: \: \: \: 1-\lambda$

Since the plane is parallel to y-axis.

$1+3\lambda =0\Rightarrow \lambda =-\frac{1}{3}$

So the equation of plane will be

$x+4z-7=0$

point $\left ( 3,2,1 \right )$ satisfies the equation.

Option 1)

$(-3,0,-1)$

Option 2)

$(-3,1,1)$

Option 3)

$(3,3,-1)$

Option 4)
$(3,2,1)$

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