 Good morning friends, I am Purva and today we will work out the following question. Find the equations of the planes that passes through three points. 1 1 0, 1 2 1 and minus 2 2 minus 1. Now let a plane passes through three points whose position vectors are vector A, vector B and vector C. Then the vector equation of the plane is given by vector R minus vector A dot product width, the cross product of vector B minus vector A and vector C minus vector A is equal to 0. And the Cartesian equation is determinant of x minus x1, y minus y1, z minus z1, x2 minus x1, y2 minus y1, z2 minus z1, x3 minus x1, y3 minus y1, z3 minus z1 is equal to 0. So this is the key idea behind a question. Let us begin with the solution now. Now we are given that the plane passes through 1 1 0, 1 2 1 and minus 2 2 minus 1. Thus the position vectors are vector A is equal to i cap plus j cap, vector B is equal to i cap plus 2 j cap plus k cap and vector C is equal to minus 2 i cap plus 2 j cap minus k cap. Now by key idea we know that the vector equation is given by vector R minus vector A dot product width, cross product of vector B minus vector A and vector C minus vector A is equal to 0. So putting the values of vector A, vector B and vector C in this equation we get vector R minus vector A that is i cap plus j cap dot product width vector B that is i cap plus 2 j cap plus k cap minus vector A that is minus i cap minus j cap cross product width vector C that is minus 2 i cap plus 2 j cap minus k cap minus vector A that is minus i cap minus j cap is equal to 0. That is we get vector R minus i cap plus j cap dot product width the cross product of. Now here i cap and i cap will cancel out 2 j cap minus j cap gives j cap plus k cap and here again minus 2 i cap minus i cap gives minus 3 i cap 2 j cap minus j cap gives plus j cap minus k cap is equal to 0. Now the Cartesian equation is given by determinant of x minus x1 y minus y1 z minus z1 x2 minus x1 y2 minus y1 z2 minus z1 x3 minus x1 y3 minus y1 z3 minus z1 is equal to 0 that is we have determinant of x minus now here x1 y1 and z1 are 1 1 and 0 so we get x minus 1 y minus 1 z minus 0 here x2 y2 and z2 are 1 2 and 1 so we get 1 minus 1 2 minus 1 1 minus 0 and here x3 y3 and z3 are minus 2 2 and minus 1 so we get minus 2 minus 1 2 minus 1 minus 1 minus 0 this is equal to 0. Now expanding along the first row we get x minus 1 into 2 minus 1 is 1 1 into minus 1 is minus 1 minus 2 minus 1 is 1 1 into 1 is 1 minus y minus 1 into 1 minus 1 is 0 and 0 multiplied by anything gives 0 minus minus 2 minus 1 is minus 3 minus 3 into 1 is minus 3 plus z minus 0 into 1 minus 1 is 0 again 0 multiplied by anything gives 0 minus minus 2 minus 1 is minus 3 minus 3 into 1 is minus 3 and this is equal to 0 that is we have x minus 1 into minus 2 minus y minus 1 into 3 plus z into 3 is equal to 0 that is we have minus 2x plus 2 minus 2 multiplied by x gives minus 2x minus 2 into minus 1 gives minus 3y plus 3 plus 3z is equal to 0 that is we have minus 2x minus 3y plus 3z plus 5 is equal to 0 or we can write this as 2x plus 3y minus 3z minus 5 is equal to 0 thus we get our answer as 2x plus 3y minus 3z is equal to 5 hope you have understood the solution bye and take care