 Hello friends and how are you all doing today? The question says using the properties of determinants show that the value of this determinant is equal to a square bracket a plus x plus y plus z. Let's proceed with the solution. Now from the left hand side we will proceed. Let d be equal to determinant a plus x x x y a plus y y z z a plus z. Now let us operate in column one. Let us add all the three columns. So on doing so we have d equal to a plus x plus y plus z a plus x plus y plus z a plus x plus y plus z in the first column and the remaining will be same in the other columns. Now on taking out a plus x plus y plus z common we are left with 1 1 1 in the first column y a plus y y z z a plus z the remaining columns. Now in row 1 let us subtract row 2 from row 1. On doing so we have d equal to a plus x plus y plus z here it will be 0. Here it will be y minus a minus y that gives us minus a here z minus z 0. The remaining will be same further now let us operate in row 2 let us subtract row 3 from it and on doing so we get d equal to a plus x plus y plus z. Now here we will have 0 minus a 0 now again 0 a minus z and here now we have third row remains the same. Right now on expanding C1 we get a plus x plus y plus z into 0 minus 0 plus 1 into a square minus 0. That further gives us the right hand side which will now be a plus x plus y plus z into a square 1 into a square that is our right hand side. Right so this completes the session hope you understood the solution well and enjoyed it too. Right so hence we have shown the question.