Question

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Answer 1

Answer:

hope it helps...

Equation of the plane y-3z+6 =0

Step-by-step explanation:

Equation of the plane passing through the line of intersection of the planes r.(i + j + k) = 1 and r.(2i + 3j - k) + 4 = 0

are given by P1+λP2 ,here λ is a real number ,can be calcultaed by given conditions

1) Convert the given plane in Cartesian form

let

Now put these equation of planes in P1+λP2

---eq1

since plane is parallel to x-axis,and Direction ratio of x-axis are (1,0,0)

and we also know that in equation of plane Dr's of normal to the plane are given,these are(1+2λ ,1+3λ,1-λ)

Because normal to the plane and x axis are perpendicular, so apply the condition of perpendicularity

(1+2λ)(1)+(0)(1+3λ)(0)(1-λ)=0

1+2λ=0

λ=-1/2

Now put the value of λ in the  eq1

Is the required equation of the plane.