The point to which origin is shifted in order to remove the first degree terms in\[2{x^2} + 5xy + 3{y^2} + 6x + 7y + 1 = 0\].

Hint: In such types of questions assume a point and new coordinates. Put it in the given equation, arrange it in descending order of powers. As per your question, equate the unnecessary terms to 0. You will end up with the required points.

Complete step-by-step answer:
Let (x,y) and (X,Y) are the coordinates in the old and new coordinate system respectively and the origin be shifted to point (h,k).
\[ \Rightarrow x = X + h;y = Y + h\]
So, the new transformed equation will be;
\[
   \Rightarrow 2{x^2} + 5xy + 3{y^2} + 6x + 7y + 1 = 0 \\
   \Rightarrow 2{(X + h)^2} + 5(X + h)(Y + k) + 3{(Y + k)^2} + 6(X + h) + 7(Y + k) + 1 = 0 \\
   \Rightarrow 2({X^2} + {h^2} + 2hX) + 5(XY + kX + hY + kh) + 3({Y^2} + {k^2} + 2kY) + 6X + 7Y + 6h + 7k + 1 = 0 \\
   \Rightarrow (2{X^2} + 5XY + 3{Y^2}) + X(4h + 5k + 6) + Y(5h + 6k + 7) + ({h^2} + {k^2} + 6h + 7k + hk + 1) = 0 \\
\]
Now, we have to remove the first degree terms.
\[ \Rightarrow \]\[4h + 5k + 6 = 0\] and \[5h + 6k + 7 = 0.\]
Solving both equations;
\[
  4h + 5k = - 6......Eq01 \\
  5h + 6k = - 7......Eq02 \\
\]
Multiplying Eq.01 by 5 and Eq.02 by 4 and then subtract;
20h+25k-(20h+24k)=-30+28
On solving we get k=-2 and then putting the value of k in Eq.01, we get h=1.
So, the origin must be shifted toward (1,-2) to remove the first degree terms.

Note: Choose the variable for your point and coordinate system wisely as there are lengthy calculations in the question. The shifted origin has the coordinates (h, k). That is, the shifted X and Y axes are at distances h and k from the original X and Y axes respectively.