To find the gradient of a function, we need to take the partial derivative with respect to each variable. In this case, we are given the function t = x^2y^e^z and the point P(1, 5, -2).

The gradient of a function is a vector that points in the direction of the steepest increase of the function at a given point. It is calculated by taking the partial derivative of the function with respect to each variable and putting them together as a vector.

Partial derivative with respect to x:
To find the partial derivative of t with respect to x, we treat y and z as constants and differentiate x^2 with respect to x. The derivative of x^2 with respect to x is 2x. Therefore, ∂t/∂x = 2xy^e^z.

Partial derivative with respect to y:
To find the partial derivative of t with respect to y, we treat x and z as constants and differentiate y^e^z with respect to y. The derivative of y^e^z with respect to y is e^z * y^(e^z - 1). Therefore, ∂t/∂y = x^2 * e^z * y^(e^z - 1).

Partial derivative with respect to z:
To find the partial derivative of t with respect to z, we treat x and y as constants and differentiate y^e^z with respect to z. The derivative of y^e^z with respect to z is e^z * ln(y) * y^e^z. Therefore, ∂t/∂z = x^2 * e^z * ln(y) * y^e^z.

Putting it all together:
The gradient of t at the point P(1, 5, -2) is a vector formed by the partial derivatives evaluated at that point. Therefore, the gradient of t at P(1, 5, -2) is given by:
∇t = (2xy^e^z, x^2 * e^z * y^(e^z - 1), x^2 * e^z * ln(y) * y^e^z)

Substituting the values x = 1, y = 5, and z = -2 into the gradient vector, we get:
∇t = (2*1*5^e^-2, 1^2 * e^-2 * 5^(e^-2 - 1), 1^2 * e^-2 * ln(5) * 5^e^-2)

Simplifying the expression further, we have:
∇t = (10e^-2, e^-2 * 5^(e^-2 - 1), e^-2 * ln(5) * 5^e^-2)

Comparing this result with the given options, we can see that option B, which is 10i + j + 0.135k, matches the calculated gradient vector. Therefore, the correct answer is option B.