 Good morning friends. I am Poojwa and today we will discuss the following question. Find the equations of the plane that passes through three points. 1, 1, minus 1, 6, 4, minus 5 and minus 4, minus 2, 3. 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 with vector b minus vector a cross product with vector c minus vector a is equal to 0 and 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. So this is the key idea behind a question. Let us begin with the solution now. Now we are given that plane passes through the points 1, 1, minus 1, 6, 4, minus 5 and minus 4, minus 2, 3. Thus the position vectors are vector a is equal to 1 i cap plus 1 j cap minus 1 k cap that is this is equal to i cap plus j cap minus k cap vector b is equal to 6 i cap plus 4 j cap minus 5 k cap and vector c is equal to minus 4 i cap minus 2 j cap plus 3 k cap. Now the vector equation of the plane is given by vector r minus vector a dot product with vector b minus vector a cross product with vector c minus vector a is equal to 0. We mark this as equation 1. Now putting the values of vector a vector b and vector c in equation 1 we get vector r minus vector a which is equal to i cap plus j cap minus k cap dot product with vector b minus vector a that is 6 i cap plus 4 j cap minus 5 k cap minus vector a that is i cap plus j cap minus k cap cross product with vector c minus vector a vector c is minus 4 i cap minus 2 j cap plus 3 k cap minus vector a is i cap plus j cap minus k cap and this is equal to 0 or we can write this as r cap minus i cap plus j cap minus k cap dot product with now 6 i cap minus i cap plus gives 5 i cap 4 j cap minus j cap gives plus 3 j cap and minus 5 k cap minus into minus becomes plus so plus k cap gives minus 4 k cap cross product with minus 4 i cap minus i cap gives minus 5 i cap minus 2 j cap minus j cap gives minus 3 j cap and 3 k cap minus into minus becomes plus so plus k cap gives plus 4 k cap and this is equal to 0. Now here we see that the cross product of vector b minus vector a and vector c minus vector a is equal to 0 as vector b minus vector a is equal to minus of vector c minus vector a so we get the points of collinear and there will be infinite number of planes passing through the given points so as we have already got that infinite number of planes passes through the given points so here we need to find the Cartesian equation of the plane so we get this as our answer hope you have understood the solution bye and take care.