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Find the distance of the point (2, 12, 5) from the point of intersection of the line

`vecr=2hati-4hat+2hatk+lambda(3hati+4hatj+2hatk) `

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#### Solution

Let the point of intersection of the line and the plane be(x_{1},y_{1},z_{1})

Equation of line is `(x−2)/3=(y+4)/4=(z−2)/2` Point (x_{1},y_{1},z_{1}) will satisfy the equation of the line.

`(x_1−2)/3=(y_1+4)/4=(z_1−2)/k`

`⇒x_1=3k+2; y_1=4k−4; z_1=2k+2 ..........(1)`

Equation of plane is *x-2y+z=0*.

Point (x_{1},y_{1},z_{1}) will also satisfy the equation of plane.

`⇒x_1−2y_1+z_1=0 `

Substituting equation (1) in equation (2), we get:

3k+2−2(4k−4)+2k+2=0

⇒k=4

Hence,

`x_1=3(4)+2;`

` y_1=4(4)−4;`

` z_1=2(4)+2`

` x_1=14;`

`y_1=12; `

`z_1=10 `

Thus, the point of intersection is (14,12,10).

Distance between the point (14,12,10) and (2,12,5)`=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)`

`=sqrt((14-2)^2+(12-12)^2+(10-5)^2)`

`=sqrt(144+0+25)`

`=sqrt169`

`=13 units`

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